Đáp án:
$\begin{array}{l}
c)\sin \left( {\pi .sinx} \right) = 1\\
\Rightarrow \pi .\sin x = \dfrac{\pi }{2} + k2\pi \\
\Rightarrow \sin x = \dfrac{1}{2} + 2k\\
Do: - 1 \le \sin x \le 1\\
\Rightarrow k = 0\\
\Rightarrow \sin x = \dfrac{1}{2}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \pi - \dfrac{\pi }{6} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\\
d)\tan \left( {\dfrac{\pi }{4}.\cos x} \right) = 1\\
\Rightarrow \dfrac{\pi }{4}.\cos x = \dfrac{\pi }{4} + k\pi \\
\Rightarrow \cos x = 1 + \dfrac{k}{4}\\
Do: - 1 \le \cos x \le 1\\
\Rightarrow k = 0\\
\Rightarrow \cos x = 1\\
\Rightarrow x = k2\pi
\end{array}$
