Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
\sin \left( {\pi .\cos x} \right) = 0\\
\Leftrightarrow \pi .\cos x = k\pi \,\,\,\,\left( {k \in Z} \right)\\
- 1 \le \cos x \le 1 \Rightarrow - \pi \le \pi .\cos x \le \pi \\
\Rightarrow \left[ \begin{array}{l}
\pi .\cos x = - \pi \\
\pi .\cos x = 0\\
\pi .\cos x = \pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = - 1\\
\cos x = 0\\
\cos x = 1
\end{array} \right. \Leftrightarrow x = \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
h,\\
\cos \left( {\pi .\sin x} \right) = - 1\\
\Leftrightarrow \pi .\sin x = \pi + k2\pi \,\,\,\,\,\left( {k \in Z} \right)\\
- 1 \le \sin x \le 1 \Leftrightarrow - \pi \le \pi .\sin x \le \pi \\
\Rightarrow \left[ \begin{array}{l}
\pi .\sin x = \pi \\
\pi .\sin x = - \pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = - 1
\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
k,\\
\sin x = \sin \left( {2x - \dfrac{\pi }{3}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2x - \dfrac{\pi }{3} + k2\pi \\
x = \pi - \left( {2x - \dfrac{\pi }{3}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- x = - \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{{4\pi }}{3} - 2x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{{4\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
l,\\
\tan x = \tan 4x\\
\Leftrightarrow x = 4x + k\pi \\
\Leftrightarrow - 3x = k\pi \\
\Leftrightarrow x = \dfrac{{k\pi }}{3}\,\,\,\left( {k \in Z} \right)\\
m,\\
\sin x = \cos 3x\\
\Leftrightarrow \cos \left( {\dfrac{\pi }{2} - x} \right) = \cos 3x\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{2} - x = 3x + k2\pi \\
\dfrac{\pi }{2} - x = - 3x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k2\pi \\
2x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
n,\\
\tan x.\tan 3x = 1\\
\Leftrightarrow \tan x = \dfrac{1}{{\tan 3x}}\\
\Leftrightarrow \tan x = \cot 3x\\
\Leftrightarrow \tan x = \tan \left( {\dfrac{\pi }{2} - 3x} \right)\\
\Leftrightarrow x = \dfrac{\pi }{2} - 3x + k\pi \\
\Leftrightarrow 4x = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
