Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\sin \left( {\dfrac{\pi }{3}\sin x} \right) = 1\\
\Leftrightarrow \dfrac{\pi }{3}\sin x = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow \sin x = \dfrac{3}{2} + 6k\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
- 1 \le \sin x \le 1 \Leftrightarrow - 1 \le \dfrac{3}{2} + 6k \le 1 \Leftrightarrow - \dfrac{5}{{12}} \le k \le - \dfrac{1}{{12}}\\
k \in Z \Rightarrow ptvn\\
2,\\
\sin \left( {\dfrac{\pi }{4}\cos x} \right) = 0\\
\Leftrightarrow \dfrac{\pi }{4}.\cos x = k\pi \\
\Leftrightarrow \cos x = 4k\\
- 1 \le \cos x \le 1 \Rightarrow - 1 \le 4k \le 1 \Leftrightarrow - \dfrac{1}{4} \le k \le \dfrac{1}{4}\\
k \in Z \Rightarrow k = 0 \Rightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
3,\\
\cos \left( {\dfrac{\pi }{5}.\sin 3x} \right) = 1\\
\Leftrightarrow \dfrac{\pi }{5}.\sin 3x = k2\pi \\
\Leftrightarrow \sin 3x = 10k\\
- 1 \le \sin 3x \le 1 \Rightarrow - 1 \le 10k \le 1 \Leftrightarrow - \dfrac{1}{{10}} \le k \le \dfrac{1}{{10}}\\
k \in Z \Rightarrow k = 0 \Rightarrow \sin 3x = 0 \Leftrightarrow 3x = k\pi \Leftrightarrow x = \dfrac{{k\pi }}{3}\\
4,\\
\cos \left( { - \dfrac{\pi }{6}.\cos 2x} \right) = 0\\
\Leftrightarrow - \dfrac{\pi }{6}.\cos 2x = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow \cos 2x = - 3 + 6k\\
- 1 \le \cos 2x \le 1 \Leftrightarrow - 1 \le - 3 + 6k \le 1 \Leftrightarrow \dfrac{1}{3} \le k \le \dfrac{2}{3}\\
k \in Z \Rightarrow ptvn
\end{array}\)
