Exercício sobre trigonometria

Resolver as equações:

$\,sen\,2x\,=\,\dfrac{\,1\,}{\,2\,}\,$

$\,sen\,3x\,=\,\dfrac{\sqrt{\,2\,}}{2}\,$

$\,sen\,(x\,-\,\dfrac{\pi}{3})\,=\,\dfrac{\,\sqrt{\,3\,}\,}{2}\,$

$\,sen\,2x\,=\,sen\,x\,$

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 $x\,=\,\frac{\pi}{3}\,+\, \frac{2k\pi}{3}\rbrace\,$
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