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 $\;x\,=\,\dfrac{5\pi}{6}\,+\,2k\pi,\,k\,\in\,\mathbb{Z}\rbrace\,$
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