Exercício sobre trigonometria

Resolver as equações a seguir:

$\phantom{X}cos\,2x\,=\,\dfrac{\,\sqrt{\,3\,}\,}{\,2\,}\phantom{X}$

$\phantom{X}cos\,2x\,=\,cos\,x\phantom{X}$

$\phantom{X}cos\left(x\,+\,\dfrac{\,\pi\,}{\,6\,}\right) = 0\phantom{X}$

$\phantom{X}cos\left(x\,-\,\dfrac{\,\pi\,}{\,4\,}\right)\,=\,1\phantom{X}$

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) $\,S\,=\,\lbrace\,x\,\in\,{\rm\,I\!R}\;|\;\,x\,=\,\frac{\pi}{4}\,+\,2k\pi \rbrace\,$

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