Lista de exercícios do ensino médio para impressão

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Um prisma triangular regular tem as arestas da base medindo 5 cm e 7 cm . Calcular a área da base, a área lateral, a área total e o volume.

resposta: $\phantom{X}A_{base}\,=\,\frac{25\sqrt{\,3\,}}{4}\;$ cm² $\phantom{X}A_{lateral}\,=\,105\;$ cm² $\phantom{X}A_{t}\,=\,\frac{5(42+5\sqrt{\,2\,}}{2}\;$ cm²$\phantom{X}V\,=\,\frac{175\sqrt{\,3\,}}{2}\;$ cm³
×
(FEI - 1977) Para quais valores de $\,p\,$ a equação $\phantom{X}tg\;px\,=\,cotg\;px\phantom{X}$ tem $\,x\,=\,\dfrac{\,\pi\,}{\,2\,}\,$ para raiz?

resposta: $\,p = \frac{1}{2}\,+\,k,\,k\,\in\,\mathbb{Z}\,$

×
(MAPOFEI - 1975) Resolver a equação $\phantom{X}cotg\;x\;-\;sen\;2x\;=\;0\phantom{X}$.

resposta: $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\pm\frac{\pi}{4}\,+\,k\pi\,\rbrace\,$

×
Resolver as equações:
a)
$\phantom{X}sec^2\,x\,=\,2\,\centerdot\,tg\,x\phantom{X}$
b)
$\phantom{X}\dfrac{1}{sen^2\,x}\,=\,1\,-\,\dfrac{cos\,x}{sen\,x}\phantom{X}$
c)
$\phantom{X}sen\,2x\,\centerdot\,cos(x\,+\,\dfrac{\pi}{\,4\,})\,=\,cox\,2x\,\centerdot\,sen(x\,+\,\dfrac{\,\pi\,}{4})\phantom{X}$
d)
$\phantom{X}(1\,-\,tg\,x)(1\,+\,sen\,2x)\,=\,1\,+\,tg\,x\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{4}\,+\,k\pi\,\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{2}\,+\,k\pi\;$ ou $\;x\,=\,\frac{3\pi}{4}\,+\,k\pi\rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{4}\,+\,k\pi\;\rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{3\pi}{4}\,+\,k\pi\;$ ou $\;x\,=\,k\pi\rbrace\,$

×
Resolver as equações:
a)
$\phantom{X}tg\,x\,=\,tg\,\dfrac{\,\pi\,}{5}\phantom{X}$
b)
$\phantom{X}cotg\,x\,=\,cotg\,\dfrac{\,5\pi\,}{6}\phantom{X}$
c)
$\phantom{X}3\,\centerdot\,tg\,x\,=\,\sqrt{\,3\,}\phantom{X}$
d)
$\phantom{X}cotg\,x\,=\,0\phantom{X}$
e)
$\phantom{X}cotg\,x\,=\,-1\phantom{X}$
f)
$\phantom{X}tg\,3x\,-\,tg\,2x\,=\,0\phantom{X}$
g)
$\phantom{X}tg\,2x\,=\,tg\,(x\,+\,\dfrac{\pi}{4})\phantom{X}$
h)
$\phantom{X}tg\,4x\,=\,1\phantom{X}$
i)
$\phantom{X}cotg\,2x\,=\,cotg(x\,+\,\dfrac{\pi}{4})\phantom{X}$
j)
$\phantom{X}tg^2\,2x\,=\,3\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{5}\,+\,k\pi\,\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{5\pi}{6}\,+\,k\pi\;\rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{6}\,+\,k\pi\;\rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{2}\,+\,k\pi\;\rbrace\,$
e) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{3\pi}{4}\,+\,k\pi\;\rbrace\,$
f) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,k\pi\;\rbrace\,$
g) $\,S\,=\,\varnothing\; \Leftarrow$
h) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{16}\,+\,\frac{k\pi}{4}\;\rbrace\,$
i) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{4}\,+\,k\pi\;\rbrace\,$
j) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{6}\,+\,\dfrac{k\pi}{2}\;$ ou $\phantom{X}x = \frac{\pi}{3} + \frac{k\pi}{2}\rbrace\,$

×
Resolver as equações:
a)
$\phantom{X}sen\,x\,\,-\,\sqrt{\,3\,}\,\centerdot\,cosx\,=\,0\phantom{X}$
b)
$\phantom{X}sen^2\,x\,=\,cos^2\,x\phantom{X}$
c)
$\phantom{X}tg\,x\,+\,cotg\,x\,=\,2\phantom{X}$
d)
$\phantom{X}sec^2\,x\,=\,1\;+\;tg\,x\,\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{3}\,+\,k\pi\,\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{4}\,+\,k\pi\;ou\;x\,=\,\frac{3\pi}{4}\,+\,k\pi;\rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{4}\,+\,k\pi\,\rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,k\pi\;ou\;x\,=\,\frac{\pi}{4}\,+\,k\pi\;\rbrace\,$

×
Resolver as equações:
a)
$\phantom{X}tg\,x\,=\,1\phantom{X}$
b)
$\phantom{X}cotg\,x\,=\,\sqrt{\,3\,}\phantom{X}$
c)
$\phantom{X}tg\,x\,=\,-\sqrt{\,3\,}\phantom{X}$
d)
$\phantom{X}tg\,x\,=\,0\phantom{X}$
f)
$\phantom{X}tg\,2x\,=\,\sqrt{\,3\,}\phantom{X}$
g)
$\phantom{X}tg\,2x\,=\,tg\,x\phantom{X}$
h)
$\phantom{X}tg\,3x\,=\,1\phantom{X}$
i)
$\phantom{X}tg\,5x\,=\,tg\,3x\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{4}\,+\,k\pi\,\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{6}\,+\,k\pi\,\rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{2\pi}{3}\,+\,k\pi\,\rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,k\pi\,\rbrace\,$
e) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{6}\,+\,\frac{k\pi}{2}\,\rbrace\,$
f) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,k\pi\,\rbrace\;$
g) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{12}\,+\,\frac{k\pi}{3}\,\rbrace\,$
h) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{k\pi}{2}, k é par\,\rbrace\,$

×
(MAUÁ - 1977) Dada a equação $\phantom{X}(sen\,x\,+\,cos\,y)(sec\,x\,+\,cossec\,y)\,=\,4\phantom{X}$:

a) resolva-a se $\phantom{X}x\,=\,y\phantom{X}$
b) resolva-a se $\phantom{X}sen\,x\,=\,cos\,y\phantom{X}$

resposta: a) x = y = π/4 + kπ b) x = π/4 + kπ e y + x = π/2 + 2kπ
×
Determinar os ângulos internos de um triângulo ABC sabendo que $\phantom{X}cos(A\,+\,B)\,=\,\dfrac{\,1\,}{\,2\,}\phantom{X}$ e $\phantom{X}sen(B\,+\,C)\,=\,\dfrac{\,1\,}{\,2\,}\phantom{X}$

resposta: A = π/6, B = π/6 e C = 2π/3
×
Resolver as equações:
a)
$\phantom{X}cos\,3x\,-\,cos\,x\,=\,0\phantom{X}$
b)
$\phantom{X}cos\,5x\,=\,cos\left(x\,+\,\dfrac{\,\pi\,}{\,3\,}\right)\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,k\pi\;ou\;x\,=\,\,\frac{k\pi}{2}\,\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\frac{\pi}{12} + \frac{k\pi}{2} \; ou\; x\,=\frac{-\pi}{18}\,+\,\frac{k\pi}{3} \rbrace\,$

×
Resolver as equações a seguir:
a)
$\phantom{X}cos\,2x\,=\,\dfrac{\,\sqrt{\,3\,}\,}{\,2\,}\phantom{X}$
b)
$\phantom{X}cos\,2x\,=\,cos\,x\phantom{X}$
c)
$\phantom{X}cos\left(x\,+\,\dfrac{\,\pi\,}{\,6\,}\right) = 0\phantom{X}$
d)
$\phantom{X}cos\left(x\,-\,\dfrac{\,\pi\,}{\,4\,}\right)\,=\,1\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{\pi}{12}\,+\,k\pi\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,2k\pi \; ou\; x\,=\frac{2k\pi}{3}\, \rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\frac{\pi}{3}\,+\,2k\pi\;ou\;,x\,=\,-\frac{2\pi}{3}\,+\,2k\pi \rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\frac{\pi}{4}\,+\,2k\pi \rbrace\,$

×
Resolver as seguintes equações:
a)
$\phantom{X}cos x = -\dfrac{ 1 }{ 2 }\phantom{X}$
b)
$\phantom{X}cos\,x\,=\,-\,\dfrac{\,\sqrt{\,2\,}\,}{\,2\,}\phantom{X}$
c)
$\phantom{X}cos\,x\,=\,\dfrac{\,\sqrt{\,3\,}\,}{\,2\,}\phantom{X}$
d)
$\phantom{X}sec\,x\,=\,\,2\phantom{X}$
e)
$\phantom{X}2\,\centerdot\,cos^2\,x\,=\,cos\,x\phantom{X}$
f)
$\phantom{X}4\,\centerdot\,cos\,x\,+\,3\,sec\,x\,=\,8\phantom{X}$
g)
$\phantom{X}2\,-\,2\,\centerdot\,cos\,x\,=\,sen\,x\,\centerdot\,tg\,x\phantom{X}$
h)
$\phantom{X}2\,\centerdot\,sen^2\,x\,+\,6\,\centerdot\,cos\,x\,=\,5\,+\,cos\,2x\phantom{X}$
i)
$\phantom{X}1\,+\,3\,\centerdot\,tg^2\,x\,=\,5\,\centerdot\,sec\,x\phantom{X}$
j)
$\phantom{X}\left(\,4\,-\,\dfrac{3}{sen^2x}\right)\,\centerdot\,\left(\,4\,-\,\dfrac{1}{cos^2x}\,\right)\,=\,0\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{2\pi}{3}\,+\,2k\pi\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\frac{3\pi}{4}\,+\,2k\pi\, \rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\frac{\pi}{6}\,+\,2k\pi \rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\frac{\pi}{3}\,+\,2k\pi \rbrace\,$
e) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\frac{\pi}{3}\,+\,2k\pi\,ou\,x\,=\,\frac{\pi}{2}\,+\,k\pi\,\rbrace\,$
f) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\frac{\pi}{3}\,+\,2k\pi \rbrace\,$
g) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,2k\pi \rbrace\,$
h) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,2k\pi\,ou\,x\,=\,\pm\frac{\pi}{3}\,+\,2k\pi \rbrace\,$
i) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\frac{\pi}{3}\,+\,2k\pi \rbrace\,$
j) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\frac{\pi}{3}\,+\,2k\pi \; ou \; x\,=\,\pm\frac{2\pi}{3}\,+\,2k\pi\rbrace\,$

×
Resolver as equações:
a)
$\phantom{X}4 \centerdot cos^2 x = 3\phantom{X}$
b)
$\phantom{X}cos^2\,x\,+\,cos\,x\,=\,0\phantom{X}$
c)
$\phantom{X}sen^2\,x\,=\,1\,+\,cos\,x\phantom{X}$
d)
$\phantom{X}cos\,2x\,+\,3\,\centerdot\,cos\,x\,+\,2\,=\,0\phantom{X}$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{\pi}{6}\,+\,2k\pi \, ou \, x\,=\,\pm\,\frac{5\pi}{6}\,+\,2k\pi\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\frac{\pi}{2}\,+\,k\pi\;ou\;x\,=\,\pi\,+\,2k\pi \, \rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\frac{\pi}{2}\,+\,k\pi\;ou\;x\,=\,\pi\,+\,2k\pi \, \rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pi + 2k\pi \; ou \; x\,=\,\pm\,\frac{2\pi}{3}\,+\,2k\pi \rbrace\,$

×
Resolver as equações:
a)
$\,cos\,x\,=\,cos\,\dfrac{\,\pi\,}{5}\,$
b)
$\,sec\,x\,=\,sec\,\dfrac{\,2\pi\,}{3}\,$
c)
$\,cos\,x\,=\,0\,$
d)
$\,cos\,x\,=\,1\,$
e)
$\,cos\,x\,=\,-1\,$
f)
$\,cos\,x\,=\,\dfrac{\,1\,}{2}\,$
g)
$\,cos\,x\,=\,\dfrac{\,\sqrt{\,2\,}\,}{2}\,$
h)
$\,cos\,x\,=\,\dfrac{\,\sqrt{\,3\,}\,}{2}\,$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{\pi}{5}\,+\,2k\pi \, \rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{2\pi}{3}\,+\,2k\pi \, \rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{\pi}{2}\,+\,2k\pi \, \rbrace\, =\,$ $\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\frac{\pi}{2}\,+\,k\pi \, \rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,2k\pi \, \rbrace\,$
e) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pi\,+\,2k\pi \, \rbrace\,$
f) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{\pi}{3}\,+\,2k\pi \, \rbrace\,$
g) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{\pi}{4}\,+\,2k\pi \, \rbrace\,$
h) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\pm\,\frac{5\pi}{6}\,+\,2k\pi \, \rbrace\,$

×
(MAPOFEI - 1976) Resolver o sistema $\,\left\{\begin{array}{rcr} sen\,(x\,+\,y)\,=\,0 & \\ x\,-\,y\,=\,\pi \phantom{XXX} & \\ \end{array} \right.\,$

resposta: $\,x\,=\,\frac{\pi}{2}\,+\,\frac{k\pi}{2}\,$, $\,y\,=\,-\frac{\pi}{2}\,+\,\frac{k\pi}{2}\,$
×
Determinar os ângulos internos de um triângulo sabendo que estão em progressão aritmética e que o seno da soma do menor ângulo com o ângulo médio é $\phantom{X}\dfrac{\,\sqrt{\,3\,}\,}{\,2\,}\phantom{X}$

resposta: ângulos $\,\frac{\pi}{3},\,\frac{\pi}{3},\,\frac{\pi}{3}\,$
×
Determinar o valor de $\phantom{X}x\;,\;\,x\,\in\,{\rm I\!R}\phantom{X}$ nas seguintes igualdades:

a) $\,sen\,5x\,=\,sen\,3x\phantom{XXXXX}$ b) $\,sen\,3x\,=\,sen\,2x\,$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,k\pi\,$ ou $x\,=\,\frac{\pi}{8}\,+\,\frac{k\pi}{4} \rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,2k\pi\,$ ou $x\,=\,\frac{\pi}{5}\,+\,\frac{2k\pi}{5}\rbrace\,$
×
Resolver as equações:
a)
$\,sen\,2x\,=\,\dfrac{\,1\,}{\,2\,}\,$
b)
$\,sen\,3x\,=\,\dfrac{\sqrt{\,2\,}}{2}\,$
c)
$\,sen\,(x\,-\,\dfrac{\pi}{3})\,=\,\dfrac{\,\sqrt{\,3\,}\,}{2}\,$
d)
$\,sen\,2x\,=\,sen\,x\,$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,\frac{\pi}{12}\,+\,k\pi\,$ ou $x\,=\,\frac{5\pi}{12}\,+\,k\pi\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,\frac{\pi}{12}\,+\,\frac{2k\pi}{3}\,$ ou $x\,=\,\frac{\pi}{4}\,+\,\frac{2k\pi}{3}\rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,\frac{2\pi}{3}\,+\,2k\pi\,$ ou $x\,=\,\pi\,+\,2k\pi\rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,2k\pi\,$ ou $x\,=\,\frac{\pi}{3}\,+\, \frac{2k\pi}{3}\rbrace\,$
×
(FEFAAP - 1977) Determinar os valores de x que satisfazem a equação $\phantom{X}4\,sen^{\large\,4}\,x\,-\,11\,sen^{\large\,2}\,x\,+\,6\,=\,0\phantom{X}$

resposta: $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,\pm\,\dfrac{\pi}{3}\,+\,k\pi\,\rbrace\,$
×
Resolver as equações:
a)
$\,sen\,x\,=\,sen\,\dfrac{\,\pi\,}{7}\,$
b)
$\,cossec\,x\,=\,2\,$
c)
$\,2\,\centerdot\,sen^2\,x\,=\,1\,$
d)
$\,sen^2\,x\,=\,1\,$
e)
$\,2 \centerdot sen^2\,x\,+\,sen\,x\,-\,1\,=\,0\,$
f)
$\,2 \centerdot sen\,x\,-\,cossec\,x\,=\,1\,$
g)
$\,3 \centerdot tg\,x\,=\,2 \centerdot cos\,x\,$
h)
$\,sen\,x\,+\,cos\,2x\,=\,1\,$
i)
$\,cos^2\,x\,=\,1\,-\,sen\,x\,$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{7}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{6\pi}{7}\,+\,2k\pi\,\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{5\pi}{6}\,+\,2k\pi\,\rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\pm \frac{\pi}{4}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{3\pi}{4}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{5\pi}{4}\,+\,2k\pi\,\rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{2}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{3\pi}{2}\,+\,2k\pi\,\rbrace\,$
e) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{3\pi}{2}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{5\pi}{6}\,+\,2k\pi\,\rbrace\,$
f) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{2}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{7\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,2k\pi - \frac{\pi}{6}\,\rbrace\,$
g) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{5\pi}{6}\,+\,2k\pi\,\rbrace\,$
h) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,k\pi\;$ ou $\,x\,=\,\frac{\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{5\pi}{6}\,+\,2k\pi\,\rbrace\,$
i) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,k\pi\;$ ou $\,x\,=\,\frac{\pi}{2}\,+\,2k\pi\,\rbrace\,$
×
Resolver em $\,{\rm I\!R}\,$ as seguintes equações:
a)
$\,sen^2\,x\,=\,\dfrac{\;1\;}{\;4\;}\,$
b)
$\,sen^2\,x\;-\;sen\,x\;=\;0\,$
c)
$\,2\,\centerdot\,sen^2\,x\,-\,3\,\centerdot\,sen\,x\,+\,1\,=\,0\,$
d)
$\,2\,\centerdot\,cos^2\,x\,=\,1\,-\,sen\,x\,$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{5\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{7\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,- \frac{\pi}{6}\,+\,2k\pi\,\rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,k\,\pi\;$ ou $\,x\,=\,\frac{\pi}{2}\,+\,2k\pi\,\rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{2}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{5\pi}{6}\,+\,2k\pi\,\rbrace\,$
d)$\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\,=\,\frac{\pi}{2}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{-\pi}{6}\,+\,2k\pi\;$ ou $\,x\,=\,\frac{7\pi}{6}\,+\,2k\pi\,\rbrace\,$

×
Resolver as equações a seguir:
a)
$\,cossec\;x\;=\;cossec\;\dfrac{\,2\pi\,}{\;3\;}\,$
b)
$\,sen\;x\;=\;\dfrac{\,\sqrt{\,3\,}\,}{\;2\;}\,$
c)
$\,sen\,x\;=\;1\,$
d)
$\,sen\,x\;=\;-1\,$
e)
$\,sen\,x\;=\;\dfrac{\;1\;}{\;2\;}\,$

resposta: a) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\;=\;\frac{2\pi}{3}\,+\,2k\pi\;{\text ou }\,x\,=\,\frac{\pi}{3}\,+\,2k\pi \rbrace\,$
b) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\;=\;\frac{\pi}{3}\,+\,2k\pi\;{\text ou }\,x\,=\,\frac{2\pi}{3}\,+\,2k\pi \rbrace\,$
c) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\;=\;\frac{\pi}{2}\,+\,2k\pi \rbrace\,$
d) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\;=\;\frac{3\pi}{2}\,+\,2k\pi\,\rbrace\,$
e) $\,S\,=\,\lbrace\,x\,\in\,{\rm I\!R}\;|\;x\;=\;\frac{\pi}{6}\,+\,2k\pi\;{\text ou }\,x\,=\,\frac{5\pi}{6}\,+\,2k\pi \rbrace\,$

×
Resolver em $\,{\rm I\!R}\,$ a equação $\phantom{X}tg\,2x\;=\;1\phantom{X}$

resposta:
Devemos notar que se a tangente de 2x é 1, então $\,tg 2x = tg\,\dfrac{\,\pi\,}{\,4\,}\,$
Temos então:
$\,2x\,=\,\dfrac{\,\pi\,}{\,4\,}\,+\,k\pi\;\Rightarrow$ $\;x = \dfrac{\,\pi\,}{\,8\,}\,\,+\,\dfrac{k\pi}{2},\;k\,\in\,\mathbb{Z}\,$
O conjunto solução então:
$\,\mathbb{S}\,=\,\lbrace\,x\,\in\,{\rm I\!R}\phantom{X}|\phantom{X}x\,=\,\dfrac{\pi}{8}\,+\,\dfrac{k\pi}{2}\, ;\phantom{X} k\,\in\,\mathbb{Z}\rbrace\,$
×
Resolver em $\,{\rm I\!R}\,$ a equação $\phantom{X}cos\,2x\;=\;0\phantom{X}$

resposta:
Devemos notar que se o cosseno de 2x é zero, então $\,2x = \pm\,\dfrac{\,\pi\,}{\,2\,}\,+\,2k\pi\;\Rightarrow$ $\;x = \pm\,\dfrac{\,\pi\,}{\,4\,}\,\,+\,k\pi,\;k\,\in\,\mathbb{Z}\,$
O conjunto solução então:
$\,\mathbb{S}\,=\,\lbrace\,x\,\in\,{\rm I\!R}\phantom{X}|\phantom{X}x\,=\,\pm\,\dfrac{\pi}{4}\,+\,k\pi\,,\phantom{X} k\,\in\,\mathbb{Z}\rbrace\,$
×
Resolver em $\,{\rm I\!R}\,$ a equação $\phantom{X}cos\,x\;=\;-\,\dfrac{\;\sqrt{\,3\,}\;}{\;2\;}\phantom{X}$

resposta:
Devemos notar que $\,-\,\dfrac{\,\sqrt{\,3\,}\,}{\,2\,}\,=\,cos\,\dfrac{\,5\pi\,}{\,6}\,$, então a equação torna-se $\phantom{X}cos\,x\;=\;\,cos\,\dfrac{\,5\pi\,}{\,6}\phantom{X}$
$\,\left\{\begin{array}{rcr} x\,= \pm\,\dfrac{\,5\pi\,}{\,6\,}\,+\,2\,k\pi \\ \,k\,\in\,\mathbb{Z}\phantom{XXXX} \\ \end{array} \right.\,$
Donde obtemos o conjunto solução:
$\,\mathbb{S}\,=\,\lbrace\,x\,\in\,{\rm I\!R}\phantom{X}|\phantom{X}x\,=\,\pm\,\dfrac{5\pi}{6}\,+\,2k\pi\,,\phantom{X} k\,\in\,\mathbb{Z}\rbrace\,$
×
Resolver em $\,{\rm I\!R}\,$ a equação $\phantom{X}sen\,x\;=\;\dfrac{\;1\;}{\;2\;}\phantom{X}$

resposta:
Devemos notar que $\,\dfrac{\,1\,}{\,2\,}\,=\,sen\,\dfrac{\,\pi\,}{\,6}\,$, então a equação torna-se $\phantom{X}sen\,x\;=\;\,sen\,\dfrac{\,\pi\,}{\,6}\phantom{X}$
$\,\left\{\begin{array}{rcr} x\,= & \dfrac{\,\pi\,}{\,6\,}\,+\,2\,k\pi \phantom{XXXX} \\ ou \\ x\,= & \left(\,\pi\,-\,\dfrac{\,\pi\,}{\,6\,}\,\right)\,+\,2\,k\pi \\ \end{array} \right.\,$
$\,k\,\in\,\mathbb{Z}\,$
Donde obtemos o conjunto solução:
$\,\mathbb{S}\,=\,\lbrace\,x\,\in\,{\rm I\!R}\phantom{X}|\phantom{X}x\,=\,\dfrac{\pi}{6}\,+\,2k\pi\;$ ou $\;x\,=\,\dfrac{5\pi}{6}\,+\,2k\pi,\,k\,\in\,\mathbb{Z}\rbrace\,$
×
Resolver em $\,{\rm I\!R}\,$ a equação $\phantom{X}sen\,x\;=\;sen\,\dfrac{\;\pi\;}{\;5\;}\phantom{X}$

resposta:
1. x pode ser:
$\,x\,=\,\dfrac{\,\pi\,}{5}\,+\,2k\pi,\,k\,\in\,\mathbb{Z}\,$ ou
2. x pode ser também:
$\,x\,=\,\left(\pi\,-\,\dfrac{\,\pi\,}{5}\right)\,+\,2k\pi\,=\,$$\dfrac{\,4\pi\,}{5}\,+\,2k\pi,\,k\,\in\,\mathbb{Z}\,$
$\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,x\,=\,\frac{\pi}{5}\,+\,2k\pi\,$ $\,{\text ou}\,x\,=\,\frac{4\pi}{5}\,+\,2k\pi,\phantom{X}k\,\in\,\mathbb{Z}\rbrace\,$
×
Unindo-se as extremidades dos arcos da forma $\phantom{X}\pm \dfrac{\,\pi\,}{\,3\,}\,+\,\dfrac{\,n\pi\,}{\,2\,}\phantom{x} (n\;\in\;\mathbb{Z})\phantom{X}$ obtém-se:
a)
b)
retângulo
c)
octógono
d)
octógono regular
e)
hexágono

resposta: (C)
×
Fazer o gráfico da função $\phantom{X}f(x) = 2 \centerdot sen x\phantom{X}$ e determinar o seu período e seu conjunto Imagem.

resposta: p = 2π e Im = [-2, 2]
×
Construir o gráfico da função $\,f\,:\,{\rm I\!R}\rightarrow\,{\rm I\!R}\,$ definida por $\phantom{X}f(x) = 1 + \operatorname{cos}\left(\,2x\,-\,\dfrac{\,\pi\,}{\,4\,}\right)\phantom{X}$

resposta:

×
Determinar o conjunto domínio, o conjunto imagem e o período da função $\phantom{X}y\,=\,2\,+\,3\operatorname{cos}\left(2x\,+\,\dfrac{\,\pi\,}{\,3\,}\,\right)\phantom{X}$.

resposta: domínio: $\,\mathbb{D}\,=\,{\rm\,I\!R}\,$ - imagem: $\,Im\,=\,\left[\,-1;\,5\,\right]\,$ - período: p = π
×
(PUC PR) Se $\,f(x)\,=\,sen\,x\,,\;\,x\,\in\,{\rm\,I\!R}\,$, então:
a)
$\;0\,\lt\,f(6)\,\lt\,\dfrac{\,1\,}{\,2\,}\;$
b)
$\;-\dfrac{\,1\,}{\,2\,}\,\lt\,f(6)\,\lt\,0\;$
c)
$\;-1\,\lt\,f(6)\,\lt\,-\dfrac{\,1\,}{\,2\,}\;$
d)
$\;\dfrac{\,1\,}{\,2\,}\,\lt\,f(6)\,\lt\,-\dfrac{\,1\,}{\,2\,}\;$
e)
$\;\dfrac{\,\sqrt{\,3\;}\,}{\,2\,}\,\lt\,f(6)\,\lt\,-\dfrac{\,1\,}{\,2\,}\;$

resposta: (B)
×
Traçar o gráfico da função f(x) = 1 + sen 2x .

resposta:

×
Com relação à função $\,f:\,{\rm\,I\!R}\,\rightarrow\,{\rm\,I\!R}\,$ definida por $\phantom{X}f(x)\,=\,1\,+\,sen\,3x\phantom{X}$ forneça:

a) o conjunto imagem
b) o período

resposta: a)
O valor do seno varia entre -1 e 1, inclusive.
Então o seno de 3x também varia entre -1 e 1.
$\phantom{X}\;-1\;\leqslant\;sen\;3x\;\leqslant\;1\phantom{X}\;$
Vamos somar 1 a cada membro da expressão acima:
$\phantom{X}\;0\;\leqslant\;1\;+\;sen\;3x\;\leqslant\;2\phantom{X}$
$\phantom{X}\;0\;\leqslant\;f(x)\;\leqslant\;2\phantom{X}$
Como f(x) varia entre 0 e 2 (inclusive), o conjunto imagem é $\,Im\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\,|\,0\,\leqslant\,x\,\leqslant\,2\,\rbrace\,$ ou
Im = [0,2]
b)
Um arco 3x executa uma volta completa no ciclo trigonométrico quando o valor de 3x varia entre 0 e 2π .
$\phantom{X} 0\;\leqslant\;3x\;\leqslant\;2\pi\phantom{X}\Rightarrow$ $\phantom{X} 0\;\leqslant\;x\;\leqslant\;\dfrac{\;2\pi\;}{3}\phantom{X}$
Então um período da função inicia-se em 0 e termina em $\,\dfrac{\;2\pi\;}{3}\,$.
$\phantom{X} p\,=\,\dfrac{\;2\pi\;}{3}\,-\,0\,=\,\dfrac{\;2\pi\;}{3}\phantom{X}$
×
Calcular cos(a + b), sendo dado sen a = -3/5 e cos b = 1/3 ,
sendo que a e b estão no intervalo $\phantom{X}\left]\,\dfrac{\,3\,\pi\,}{2}\,;\,2\pi\,\right[\phantom{X}$

resposta:
Lembramos que cos(a + b) = cos a . cos b - sen a . sen b
Então:
Passo 1 - Calcular o cos a
$\,cos\,a\,=\,+\,\sqrt{\,1\,-\,sen^2\,a\,}\,=\,\dfrac{\,4\,}{5}\,$
Passo 2 - Calcular o sen b
$\,sen\,b\,=\,-\,\sqrt{\,1\,-\,cos^2\,b\,}\,=\,\dfrac{\,-\,2\,\sqrt{\,2\,}\,}{3}\,$
Passo 3 - Calcular o cos (a + b)
$\,cos\,(a\,+\,b)\,=\,cos\,a\,\centerdot\,cos\,b\,-\,sen\,a\,\centerdot\,sen\,b\,=$ $\dfrac{\,4\,}{5} \centerdot \dfrac{\,1\,}{3}\,-\,(-\dfrac{\,3\,}{5})\, \centerdot\,(-\,\dfrac{\,2\,\sqrt{\,2\,}}{3})\,=\,$ $\dfrac{4}{\,15\,} - \dfrac{\,6\,\sqrt{\,2\,}}{15}\,=$
$\boxed{\;\dfrac{\,\,4\,-\,6\sqrt{\,2\,}}{15}\;}\,$
×
Calcular sen 75°.

resposta:
Lembramos que sen(a + b) = sen a . cos b + sen b . cos a
Então:
$\,sen\,75^o\,=\,$ $\,sen\,(45^o\,+\,30^o)\,=\,$ $\,sen\,45^o\,\centerdot\,cos\,30^o\,+\,cos\,45^o\,\centerdot\,sen\,30^o\,=\,$ $\,\dfrac{\,\sqrt{\,2\,}\,}{2}\,\centerdot\,\dfrac{\,\sqrt{\,3\,}\,}{2}\,+\,\dfrac{\,\sqrt{\,2\,}\,}{2}\,\centerdot\,\dfrac{\,1\,}{2}\,\,=\,$ $\dfrac{\,\sqrt{\,6\,}\,}{4}\,+\,\dfrac{\,\sqrt{\,2\,}\,}{4}\,$
$\phantom{X}\boxed{\;sen\,75^o\,=\,\dfrac{\,\sqrt{\,6\,}\,+\,\sqrt{\,2\,}\,}{4}\,}\phantom{X}$
×
No triângulo da figura são conhecidos os ângulos Â = 60° e $\,\hat{B}\,$ = 75° e também o lado c = 13 m.

Pede-se:
a) a medida em graus do ângulo C;
b) a medida em metros dos lados a e b;

resposta:
Resolução:
a) A soma dos ângulos internos de um triângulo qualquer é igual a 180°, então $\phantom{X} \require{cancel}\hat{A}\,+\,\hat{B}\,+\,\hat{C}\,=\,180^o\;\Rightarrow$ $\;\hat{C}\,=\,180^o\,-\,(\hat{A}\,+\,\hat{B})\,=$ $\,180^o\,-\,135^o\,=\,45^o\;$

b) Pelo Teorema dos Senos temos que $\,\dfrac{b}{\,sen \hat{B}\,}\,=\,\dfrac{c}{\,sen \hat{C}\,}\,=\,\dfrac{a}{\,sen \hat{A}\,}\,$, então podemos concluir que $\,b\,=\,\dfrac{\,c\,\centerdot\,sen\,\hat{B}\,}{sen\,\hat{C}}\phantom{X}$ e $\phantom{X}a\,=\,\dfrac{\,c\,\centerdot\,sen\,\hat{A}\,}{sen\,\hat{C}}\,$
Lembrar que $\,sen(a\,+\,b)\,=$ $\,sen\,a\,\centerdot\,cos\,b\,+\,sen\,b\,\centerdot\,cos\,a\,$
$\,sen\,\hat{A}\,=\,sen75^o\,$ $=\,sen\,(45^o\,+\,30^o)\,=$ $\,sen\,45^o\,\centerdot\,sen\,30^o\,+\,sen\,30^o\,\centerdot\,sen\,45^o\,=\,$ $\dfrac{\,\sqrt{\,2\;}}{2}\,\dfrac{\,\sqrt{\,3\;}}{2} + \dfrac{\,\sqrt{\,3\;}}{2}\,\dfrac{\,\sqrt{\,2\;}}{2}\, =$ $\dfrac{\,2\sqrt{\,6\;}}{4} = \dfrac{\,\sqrt{\,6\;}}{2}$
$\,sen\,\hat{B}\,=\,sen\,60^o\,=\,\dfrac{\,\sqrt{\,3\;}}{2}\,$
$\,sen\,\hat{C}\,=\,sen45^o\,=\,\dfrac{\,\sqrt{\,2\;}}{2}\,$

$\phantom{X}a\,=\,\dfrac{\,c\,\centerdot\,sen\,\hat{A}\,}{sen\,\hat{C}}\,=$ $\,\dfrac{\,13\,\centerdot\,\dfrac{\sqrt{\,6\,}}{2}\,}{\dfrac{\sqrt{\,2\,}}{2}}\, =$ $\,13\,\centerdot\,\dfrac{\sqrt{\,6\,}\,\centerdot\,\cancel {2}}{\cancel {2}\,\centerdot\,\sqrt{\,2\,}\,}\,=\,$ $13\,\centerdot\,\dfrac{\sqrt{\,3\,}\,\centerdot\,\cancel{\sqrt{\,2\,}} }{\cancel{\sqrt{\,2\,}}\,}\,=\,13\,\sqrt{\,3\,}\, m\phantom{X}$

$\phantom{X}b\,=\,\dfrac{\,c\,\centerdot\,sen\,\hat{B}\,}{sen\,\hat{C}}\;=$ $\,\dfrac{\,13\,\centerdot\,\dfrac{\,\sqrt{\,3\;}}{2}\,}{\dfrac{\,\sqrt{\,2\;}}{2}}\,=$ $\,13\,\centerdot\,\dfrac{\sqrt{\,3\,}\,\centerdot\,\cancel {2}}{\cancel {2}\,\centerdot\,\sqrt{\,2\,}\,}\,=$ $\,13\,\centerdot\,\dfrac{\sqrt{\,3\,}\,\centerdot\,\sqrt{\,2\,}}{\sqrt{\,2\,}\,\centerdot\,\sqrt{\,2\,}\,} = \dfrac{\,13\,\sqrt{\,6\,}}{2}\; m\phantom{X}$

×
O ângulo sob o qual um observador vê uma torre duplica quando ele se aproxima 110 m e triplica quando se aproxima mais 50 m. Calcular a altura da torre.

resposta: 88 m
×
Num triângulo ABC , o ângulo Â é obtuso. Os lados AB e AC medem 3 e 4 , respectivamente. Então:
a) BC < 4
b) BC < 5
c) BC > 7
d) 5 < BC < 7
e) nenhuma das anteriores é correta

resposta: (D)
×
Exprimir 1 rad em graus. ( π ≅ 3,14 ) .

resposta: 57°19'29''
×
Exprimir 60°15' em radianos. Assuma π ≅ 3,14 .

resposta:
Resolução:
Passo 1 - converter 15 minutos em graus.
60°15' = 60° + 15' (I)
mas 1° é o mesmo que 60' , portanto fazemos uma primeira regra de três simples
$\,\left.\begin{array}{rcr} 1^o\,\longrightarrow\,60'\;& \\ x\,\longrightarrow\,15'\;& \\ \end{array} \right\}\,$ $\;\Rightarrow\,x\,=\,\dfrac{\,15'\,\centerdot\,1^o\,}{60'}\,$ $\,\Rightarrow \;x\,=\,0,25^o\;\,$
Então em (I) temos que 60°15' = 60° + 0,25°
Passo 2 - converter 60,25 graus em radianos
Sabendo que 180° é o mesmo que π radianos, fazemos uma segunda regra de três simples:
$\,\left.\begin{array}{rcr} 180^o\,\longrightarrow\,\pi\;& \\ 60,25^o\,\longrightarrow\,y\;& \\ \end{array} \right\}\,$ $\;\Rightarrow\,y\,=\,\dfrac{\,60,25^o\,\times\,3,14\,}{180^o}\,$ $\,\Rightarrow \;y\,=\,1,05\;\,$
Resposta:
$\,\;60^o15'\,\cong\,1,05\,$
×

resposta:
Resolução:
Sabendo que 180° correspondem a π radianos, escrevemos uma regra de três simples:
$\,\left.\begin{array}{rcr} 180^o\,\longrightarrow\,\pi\;& \\ 120^o\,\longrightarrow\,x\;& \\ \end{array} \right\}\,$ $\;\Rightarrow\,x\,=\,\dfrac{\,120^o\,\centerdot\,\pi\,}{180^o}\,$ $\,\Rightarrow \boxed{\;x\,=\,\dfrac{\,2\pi\,}{\;3\;}\;}\,$
Resposta:$\, \;120^o\,=\,\dfrac{\,2\pi\,}{\;3\;}\,rad\;\,$
×
(FEI MAUÁ) Calcular a distância da origem ao vértice da parábola:$\phantom{X}y\,=\,x^2\,-\,6x\,+\,10\phantom{X}$

resposta: $\,d\,=\,\sqrt{\,10\,}\,$
×
Calcular os três ângulos internos de um triângulo $\,ABC\,$ sabendo que a = 2, b = $\,\sqrt{6}\,$ e c = $\,\sqrt{3}\,$ + 1.

resposta: $\,\hat{A}\,=\,45^o,\,\hat{B}\,=\,60^o\,e\,\hat{C}\,=\,75^o\,$
×
Um triângulo tem lados a = 10 m , b = 13 m e c = 15 m . Calcular o ângulo $\,\hat{A}\,$ do triângulo.

resposta: $\,arc\,cos\dfrac{49}{65}\,$
×
(FEI - 1977) Calcular $\phantom{X}c\phantom{X}$, sabendo que:
$\,a\,=\,4\,$
$\,b\,=\,3\sqrt{\,2\,}\,$
$\,\hat{C}\,=\,45^o\,$

resposta: $\,c\,=\,\sqrt{10}\,m\,$
×
Dois lados de um triângulo medem 8 m e 12 m e formam entre si um ângulo de 120° . Calcular o terceiro lado.

resposta: $\,4\sqrt{19}\,m\,$
×
Sendo $\,x\,$ um arco do quarto quadrante, qual o sinal da expressão $\phantom{X}y\,=\,\dfrac{\,cossec\,x\,\centerdot\,cossec\,(x\,+\,\pi)\,}{\,cossec\,\left(x\,+\,\dfrac{\,\pi\,}{\,2\,}\right)\,\centerdot\,cos\,x}\phantom{X}$

resposta: negativo
×
Calcule, se existir:
a)
$\,cossec\,\dfrac{\,\pi\,}{\,2\,}\,$
b)
$\,cossec\,\dfrac{\,3\pi\,}{\,2\,}\,$
c)
$\,cossec\,\pi\,$
d)
$\,cossec\,4\pi\,$

resposta: a)1 b)-1 c)não existe d)não existe
×
Avaliar se são possíveis as seguintes igualdades: a) $\,cossec\,x\,=\,0\phantom{X}$ e b) $\,cossec\,x\,=\,2\,$

resposta: a) impossível, não existe b) possível, existe.
×
Avaliar se são possíveis as seguintes igualdades: a) $\,sec\,x\,=\,0\phantom{X}$ e b) $\,sec\,x\,=\,-2\,$

resposta: a) impossível, não existe b) possível, existe.
×
Qual é o sinal da expressão $\phantom{X}y\,=\,\dfrac{\,sec\,x\,\centerdot\,tg\,x\,\centerdot\,sen\,x\,\,}{\,cos\,\left(x\,+\,\dfrac{\,\pi\,}{\,2\,}\,\right)\,}\phantom{X}$, sendo $\,0\,\lt\,x\,\lt\,\dfrac{\,\pi\,}{\,2\,}\,$

resposta: negativo
×
Determine, se existir:
a)
$\,sec\,0\,$
b)
$\,sec\,\pi\,$
c)
$\,sec\,\dfrac{\,\pi\,}{\,2\,}\,$
d)
$\,sec\,(-\pi)\,$

resposta: a)0 b)-1 c)não existe d)-1
×
Sendo $\phantom{X}x\phantom{X}$ um arco do 3º quadrante , qual o sinal da expressão $\phantom{X}y\,=\,\dfrac{\,tg\,\left(\,x\,+\,\dfrac{\,\pi\,}{\,2\,}\,\right)\,\centerdot\,cotg\,\left(\,x\,+\,\dfrac{\,\pi\,}{\,2\,}\,\right)\,}{\,cotg\,x\,\centerdot\,cotg\,(x\,+\,\pi)\,}\phantom{X}$

resposta: positivo
×
Calcule, se existir:
a)
$\,cotg\,\dfrac{\,\pi\,}{\,2\,}\,$
b)
$\,cotg\,\dfrac{\,3\pi\,}{\,2\,}\,$
c)
$\,cotg\,4\pi\,$
d)
$\,cotg\,(-4\pi)\,$

resposta: a)0 b)0 c)não existe d)não existe
×
a) Para todo arco $\,x\,$ real, existe o arco $\phantom{X}\boxed{\; x'\,=\,\pi\,-\,x\;}\phantom{X}$ cuja imagem é simétrica à   em relação ao

b) Para todo arco $\,x \in {\rm I\!R}\,$ existe o arco $\phantom{X}\boxed{\; x'\,=\,x\,-\,\pi\,\;}\phantom{X}$ cuja imagem é simétrica à   em relação à

c) Para todo arco $\,x\,$ real, existe o arco $\phantom{X}\boxed{\; x'\,=\,2\pi\,-\,x\;}\phantom{X}$ cuja imagem é simétrica à   em relação ao

d) Para todo arco $\,x \in {\rm I\!R}\,$ existe o arco $\phantom{X}\boxed{\; x'\,=\,\dfrac{\,\pi\,}{\,2\,}\,-\,x\,\;}\phantom{X}$ cuja imagem é simétrica à   em relação à

resposta: a) imagem de $\,x\,$ - eixo dos senos
b) imagem de $\,x\,$ - origem dos eixos
c) imagem de $\,x\,$ - eixo dos cossenos
d) imagem de $\,x\,$ - reta bissetriz do primeiro quadrante

×
a)
Simplificar a expressão $\phantom{X}cos\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right)\phantom{X}$
b)
Calcular $\phantom{X}sen^2\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right) + cos^2\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right)\phantom{X}$
c)
Calcular $\phantom{X}cos^2\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right) + cos^2\,x\phantom{X}$

resposta: a)$\,sen\,x\,$ b)1 c)1
×
a)
Simplificar a expressão $\phantom{X}sen\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right)\phantom{X}$
b)
Calcular $\phantom{X}sen^2\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right) + cos^2\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right)\phantom{X}$
c)
Calcular $\phantom{X}sen^2\,x\,+\,sen^2\left(\dfrac{\,3\pi\,}{\,2\,}\,+\,x\right)\phantom{X}$

resposta: a)$\,-cos\,x\,$ b)1 c)1
×
a)
Simplificar a expressão $\phantom{X}cos\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right)\phantom{X}$
b)
Calcular $\phantom{X}sen^2\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right) + cos^2\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right)\phantom{X}$
c)
Calcular $\phantom{X}sen^2\,x\,+\,sen^2\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right)\phantom{X}$

resposta: a)$\,-cos\,x\,$ b)1 c)1
×
a)
Simplificar a expressão $\phantom{X}cos\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right)\phantom{X}$
b)
Calcular $\phantom{X}sen^2\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right) + cos^2\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right)\phantom{X}$
c)
Calcular $\phantom{X}cos^2\left(\dfrac{\,3\pi\,}{\,2\,}\,-\,x\right)\,+\,cos^2\,x\phantom{X}$

resposta: a)$\,-sen\,x\,$ b)1 c)1
×
(PUCRJ - 2018) Simplificando a expressão $\phantom{X}2\,\centerdot\,\dfrac{\,(3^6\,+\,3^5)\,}{\,3^4\,-\,3^3\,}\phantom{X}$
a)
12
b)
13
c)
3
d)
36
e)
1

resposta: (D)
×
a)
Simplificar a expressão $\phantom{X}sen\,(\frac{\,\pi\,}{\,2\,}\,+\,x)\phantom{X}$
b)
Calcular $\phantom{X}sen^2\,\frac{\,4\pi\,}{\,9\,}\,+\,cos^2\,\frac{\,4\pi\,}{\,9\,}\phantom{X}$
c)
Calcular $\phantom{X}sen^2\,\frac{\,4\pi\,}{\,9\,}\,+\,cos^2\,\frac{\,5\pi\,}{\,9\,}\phantom{X}$
d)
Calcular $\phantom{X}\frac{\,\pi\,}{\,2\,}\,-\,\frac{\,4\pi\,}{\,9\,}\phantom{X}$
e)
Calcular $\phantom{X}\frac{\,5\pi\,}{\,9\,}\,-\,\frac{\,\pi\,}{\,2\,}\phantom{X}$
f)
Calcular $\phantom{X}sen^2\,\frac{\,4\pi\,}{\,9\,}\,+\,sen^2\,\frac{\,\pi\,}{\,18\,}\phantom{X}$
g)
Calcular $\phantom{X}sen^2\,\frac{\,5\pi\,}{\,9\,}\,+\,sen^2\,\frac{\,\pi\,}{\,18\,}\phantom{X}$

resposta: a)$\,cos x\,$b)1c)1d)$\,\frac{\pi}{18}\,$e)$\,\frac{\pi}{18}\,$f)1g)1
×
(VUNESP) A expressão $\phantom{X}\dfrac{\,cos^2\,\theta\,}{\;1\,-\,sen\,\theta}\phantom{X}$, com $\,sen\,\theta\,\ne\,1\,$, é igual a:
a)
$\,sen\,\theta\,$
b)
$\,sen\,\theta\,+\,1\,$
c)
$\,tg\,\theta\,\centerdot\,cos\,\theta\,$
d)
$\,1\,$
e)
$\,\frac{\,sen\,\theta\,}{\,sec\,\theta\,}\,$

resposta: (B)
×
Sabendo que $\,sen x - cos x = a\,$, calcule:
a)$\,sen\,x\,\centerdot\,cos\,x\,$
b)$\,sen^3\,x\,-\,cos^3\,x\,$

resposta: a)$\,\dfrac{1\,-\,a^2}{2}\,$
b)$\,\dfrac{3a\,-\,a^3}{2}\,$
×
Sabe-se que $\,sen\,\dfrac{\,4\pi\,}{\,9\,}\,=\,a\,$
a)
Qual o sinal de $\,a\,$? Justifique.
b)
Calcule, em função de $\,a\,$, $\,sen\,\dfrac{\,5\pi\,}{\,9\,}\,$.
c)
Calcule $\,sen\,\dfrac{\,\pi\,}{\,18\,}\;$ e $\;cos\,\dfrac{\,\pi\,}{\,18\,}\;$

resposta: a) positivo porque o arco $\,\frac{4\pi}{9}\,$ pertence ao primeiro quadrante $\,0\,\lt\,\frac{4\pi}{9}\,\lt\,\frac{\pi}{2}\,$
b)$\,a\,$
c)$\,sen\frac{\pi}{18}\,=\,\sqrt{1\,-\,a^2}\,$ e $\,cos\frac{\pi}{18}\,=\,a\,$
×
Calcule o valor de
a) $\,sen^2\,70^o\,+\,cos^2\,100^o\,$
b) $\,sen^2\,55^o\,+\,cos^2\,55^o\,$

resposta: a)1 b)1
×
Para quais valores de $\,x\,$ temos $\,tg\,x\,=\,\sqrt{\,3\,}\,$

resposta: $\,x\,=\,\frac{\;\pi\;}{3}\,+\,k\pi,\,k\,\in\,\mathbb{Z}\,$
×
Sabendo que $\phantom{X}tg\,x\,=\,3\phantom{X}$, $\phantom{X}\pi\,\lt\,x\,\lt\,\frac{\,3\pi\,}{\,2\,}\,$, calcule $\,sen\,x\;-\;cos\,x\,$.

resposta: $\,-\frac{\sqrt{10\,}}{5}\,$
×
Dado que $\,tg\,x\,=\,-2\;$, calcule o valor de $\,\dfrac{\,4\,cos\,x\,}{\,3\,sen\,x\,}\,$.

resposta: -2/3
×
Calcule o valor da expressão $\phantom{X}y\,=\,3\,\centerdot\,tg\,\dfrac{\,\pi\,}{4}\,-\,2\,\centerdot\,tg\,\dfrac{\,\pi\,}{3}\,\centerdot\,tg\,\dfrac{\,\pi\,}{6}\,-\,tg\,\dfrac{\,3\pi\,}{4}\phantom{X}$

resposta: 2
×
Se $\,x \in {\rm I\!R}\,$ é tal que $\,|sen\,x|\,=\,|cos\,x|\,$, quanto vale $\,tg\,x\,$?

resposta: $\,\pm\,1\,$
×
Com $\phantom{X}0\,\lt\,x\,\lt\,\pi\;$ e $\;sen\,x\,=\,\dfrac{\,1\,}{\,8\,}\,$, quanto vale $\,tg\,x\,$?

resposta: $\,\pm\,\frac{\sqrt{7\,}}{21}\,$
×
Sendo $\phantom{X}sen\,x\,=\,-\dfrac{\,1\,}{\,3\,}\phantom{X}$, $\phantom{X}\pi\,\lt\,x\,\lt\,\frac{\,3\pi\,}{\,2\,}\,$, obtenha $\,tg\,x\,$

resposta: $\,\frac{\sqrt{2\,}}{4}\,$
×
Dado $\phantom{X}cos\,x\,=\,\dfrac{\,1\,}{\,6\,}\phantom{X}$, $\phantom{X}0\,\lt\,x\,\lt\,\frac{\,\pi\,}{\,2\,}\,$, determine $\,tg\,x\,$

resposta: $\,\sqrt{35\,}\,$
×
Disponha em ordem crescente os números reais $\phantom{X}tg\dfrac{\,\pi\,}{\,6\,}\,$, $\,tg\dfrac{\,\pi\,}{\,3\,}\,$, $\,tg\,\pi\,$, $\,tg\dfrac{\,\pi\,}{\,4\,}\,$, $\,tg\dfrac{\,3\pi\,}{\,4\,}\,$, $\,tg\dfrac{\,2\pi\,}{\,3\,}\,$.

resposta: $\,tg\frac{\,2\pi\,}{\,3\,}\,\lt\,tg\frac{\,3\pi\,}{\,4\,}\,\lt\,tg\,\pi\,\lt\,tg\frac{\,\pi\,}{\,6\,}\,\lt\,tg\frac{\,\pi\,}{\,4\,}\,\lt\,tg\frac{\,\pi\,}{\,3\,}$
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Determine o sinal da expressão $\phantom{X}y\,=\,tg^2\,\dfrac{\,\pi\,}{\,5\,}\,\centerdot\,tg\,\dfrac{\,4\pi\,}{\,5\,}\,\centerdot\,tg\,\dfrac{\,6\pi\,}{\,5\,}\,\centerdot\,tg\,\dfrac{\,8\pi\,}{\,5\,}\phantom{X}$.

resposta: positivo
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Encontre os valores de x para os quais temos:
a)
sen x = cos x
b)
sen² x = 1

resposta: a) $\,\frac{\pi}{4}\,+\,k\pi\;,\;k\,\in\,\mathbb{Z}\,$ b) $\,\frac{\pi}{2}\,+\,k\pi\;,\;k\,\in\,\mathbb{Z}\,$
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Sabendo que $\phantom{X}6\,cos\,x\,-\,1\,=\,4\phantom{X}$, com $\,\frac{\,3\pi\,}{\,2\,}\,\lt\,x\,\lt\,2\pi\;$, obtenha $\,sen\,x\,$

resposta: $\,-\frac{\,\sqrt{\,11\,}}{\,6\,}\,$
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Se $\,cos\,x\,=\,\frac{\,2\,}{\,3\,}\phantom{X}$ e $\,x\,$ está no primeiro quadrante, determine $\,sen x\,$ e $\,sen \left(\frac{\,\pi\,}{\,2\,}\,-\,x\,\right)\,$

resposta: $\,\dfrac{\,\sqrt{\,5\,}}{\,3\,}\,$ e $\,\dfrac{\,2\,}{\,3\,}\,$
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Sendo $\,sen\,x\,=\,\frac{\,4\,}{\,5\,}\phantom{X}$ e $\phantom{X}\frac{\,\pi\,}{\,2\,}\,\lt\,x\,\lt\,\pi\,$, determine $\,cos\,x\,$.

resposta: -3/5
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Encontre o perímetro do triângulo OAB , situado no 2º quadrante do ciclo trigonométrico.

resposta: $\,\frac{\,3\,+\,\sqrt{\,3\;}}{2}\,$
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Sendo $\phantom{X}k\;\in\;\mathbb{Z}\phantom{X}$, calcule em cada caso o valor de $\,cos\,x\,$, com:
a)
$\,x\,=\,2k\pi\,$
b)
$\,x\,=\,(2k\,+\,1)\pi\,$
c)
$\,x\,=\,(2k\,-\,1)\dfrac{\,\pi\,}{2}$
d)
$\,x\,=\,\pm\dfrac{\,\pi\,}{3}\,+\,2k\pi\,$

resposta: a)1b)-1c)0d)1/2
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Escreva a expressão geral dos arcos x para os quais temos:
a) $\,cos\,x\,=\,-\dfrac{\,\sqrt{\,2\;}}{2}\,$
b) $\,cos\,x\,=\,0\,$

resposta: a) $\,x\,=\,\pm\dfrac{\,3\pi\,}{4}\,+\,2k\pi\;,\;k\,\in\,\mathbb{Z}\,$
b) $\,x\,=\,\dfrac{\,\pi\,}{2}\,+\,k\pi\;,\;k\,\in\,\mathbb{Z}\,$
×
Para que se tenha cos x = 3m - 1 , quais são os possíveis valores de m ?

resposta: $\,0\,\leqslant\,m\,\leqslant\,\frac{2}{3}\;,\,m\,\in\,{\rm\,I\!R}\;$
×
Forneça o sinal de cada uma das expressões abaixo:
a)
cos 125° - cos 124°
b)
$cos\,\dfrac{\,\pi\,}{7}\,\centerdot \,cos\,7\pi\,\centerdot\,cos\,4\pi$
c)
(1 + cos x) . (1 - cos x) , x ∈ ${\rm I\!R}$

resposta: a)negativo b)negativo c)positivo ou nulo
×
Dê o valor de:
a)
$\,cos\,\frac{\,5\pi\,}{3}\,$
b)
$\,cos\,\frac{\,7\pi\,}{6}\,$
c)
$\,cos\,\frac{\,7\pi\,}{4}\,$
d)
$\,cos\,720^o\,$
e)
$\,cos\,120^o\,$
f)
$\,cos\,\frac{\,\pi\,}{2}\,$
g)
$\,cos\,150^o\,$
h)
$\,cos\,\left(\frac{\,\pi\,}{2}\,+\,\pi \right)\,$
i)
$\,cos\,x\,-\,cos\,y\,$ sendo $\,x\,+\,y\,=\,2\pi\,$

resposta:
a)
1/2
b)
$\,\frac{-\sqrt{3}}{2}\,$
c)
$\,\frac{\sqrt{2}}{2}\,$
d)
1
e)
-1/2
f)
0
g)
$\,\frac{-\sqrt{3}}{2}\,$
h)
0
i)
0

×
Escreva a expressão geral dos $\,arcos\;x\,$ para os quais temos $\,sen\,x\,=\,\pm\,\dfrac{\,\sqrt{\,3\;}}{2}\,$

resposta: $\,x\,=\,\frac{\,\pi\,}{3}\,+\,k\pi\,$ ou $\,x\,=\,\frac{\,2\pi\,}{3}\,+\,k\pi,\,k\,\in\,\mathbb{Z}\,$
×
Sendo $\;k\,\in\,\mathbb{Z}\;$, calcule, em cada caso, o valor de $\,sen\,x\,$, com:
a)
$\,x\,=\,(2k\,+\,1)\,\centerdot\,\pi\,$
b)
$\,x\,=\,(2k\,+\,1)\dfrac{\,\pi\,}{2}\,$
c)
$\,x\,=\,\dfrac{\,\pi\,}{4}\,+\,k\pi\,$
d)
$\,x\,=\,-\dfrac{\,\pi\,}{3}\,+\,2k\pi\,$ ou $\,x\,=\,-\dfrac{\,2\pi\,}{3}\,+\,2k\pi\,$

resposta:
a)
0
b)
$\,\pm\,1\,$
c)
$\,\pm\frac{\,\sqrt{\,2\,}\,}{2}\,$
d)
$\,-\frac{\sqrt{\,3\,}}{2}\,$

×
Dê o sinal de cada uma das expressões:
a)
$\,sen\frac{\,\pi\,}{5}\,\centerdot\, sen\frac{\,\pi\,}{3}\centerdot\,sen\frac{\,3\pi\,}{5}\centerdot\,sen\frac{\,5\pi\,}{3}\,$
b)
$\,(1\,-\,sen\,x)(1\,+\,sen\,x),\phantom{X}x\,\in\,{\rm I\!R}\,$
c)
$\,sen\,111^o\;-\;sen\,110^o$

resposta: a - negativo; b-positivo ou nulo; c-negativo
×
Calcule:
a)
sen 300°
b)
sen 330°
c)
sen $\,5\pi\,$
d)
sen$\,-\frac{\,\pi\;}{\;4\;}\,$
e)
$\,\dfrac{sen\,\frac{\,\pi\,}{\,2\,}\;-\;sen\,\frac{\,\pi\,}{\,3\,}}{sen\,\frac{\,\pi\,}{\,6\,}}\,$
f)
$\,\dfrac{sen\,\frac{\,\pi\,}{\,4\,}\;-\;sen\,\frac{\,4\pi\,}{\,3\,}}{sen^2\,\frac{\,5\pi\,}{\,6\,}}\,$

resposta:
a)
$\,\frac{-\sqrt{3}}{2}\,$
b)
-1/2
c)
0
d)
$\,\frac{-\sqrt{2}}{2}\,$
e)
$\,2 - \sqrt{3}\,$
f)
$\,-\sqrt{\,6\;}\,$

×
Associe os valores da 2ª coluna correspondentes senos da 1ª coluna:
A.
sen 120°
B.
sen$\,\dfrac{\,3\pi\;}{\;4\;}\,$
C.
sen$\,\pi\,$
D.
sen$\,\dfrac{\,4\pi\,}{\;3\;}\,$
E.
sen$\,\dfrac{\,7\pi\,}{\;4\;}\,$
F.
sen 270°
1.
$\,\dfrac{\,\sqrt{\,2\;}\,}{2}\,$
2.
-1
3.
$\,\dfrac{\,-\sqrt{\,2\;}\,}{2}\,$
4.
sen 0°
5.
$\,-\dfrac{\,\sqrt{\,3\;}\,}{2}\,$
6.
$\,\dfrac{\,\sqrt{\,3\;}\,}{2}\,$

resposta: A-6; B-1; C-4; D-5; E-3; F-2
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Calcular a 4a determinação negativa do arco de 810o

resposta: -1350o
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Calcular a terceira determinação positiva do arco de 1910o.

resposta: 830o
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Calcular a primeira determinação positiva (a0) dos seguintes arcos:
a)
1620o
b)
$\,125\dfrac{\;\pi\;}{11}\,$
c)
-810o
d)
$\,-97\dfrac{\;\pi\;}{7}\,$

resposta:
a)
ao = 180o
b)
ao=$\,15\frac{\pi}{\,11\,}\,$
c)
ao=270o
d)
ao=$\,\frac{\pi}{\,7\,}\,$

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Para os itens seguintes, considere as funções $\,f\,:\,{\rm I\!R}\,\rightarrow\,{\rm I\!R}\phantom{X}$ e $\phantom{X}g\,:\,{\rm I\!R}\,\rightarrow\,{\rm I\!R}\;$ a seguir:
$\,\left\{\begin{array}{rcr} f(x)\,=\,-\frac{\,2\,}{3}x\,+\,\frac{\,8\,}{3}\,& \\ g(x)\,=\,x\,+\,1\phantom{XXX}& \\ \end{array} \right.\,$

a) Represente num mesmo plano
cartesiano as funções f(x) e g(x) .

b) Calcule para quais valores
de $\;x\;$ as imagens
de f(x) e g(x) são iguais.
c)
Calcule para quais intervalos do $\;x\;$ temos f(x) < 0 .

d)
Calcule para quais intervalos do eixo $\;x\;$ temos g(x) < 0 .

e)
Calcule para quais intervalos do eixo $\;x\;$ temos f(x) > g(x) .

f)
Calcule os zeros das funções citadas.

resposta: a)

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Veja exercÍcio sobre: geometria métrica espacial