Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{6} + n2\pi\\x = \dfrac{5\pi}{6} + n2\pi\\x = n\dfrac{\pi}{2}\end{array}\right.\quad (n \in \Bbb Z)$
Giải thích các bước giải:
$\cos(3\pi\sin x) = \cos(\pi\sin x)$
$\Leftrightarrow 4\cos^3(\pi\sin x) - 3\cos(\pi\sin x) = \cos(\pi\sin x)$
$\Leftrightarrow \cos(\pi\sin x)[\cos^2(\pi\sin x) - 1] = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos(\pi\sin x) = 0\\\cos(\pi\sin x) = 1\\\cos(\pi\sin x) = -1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\pi\sin x = \dfrac{\pi}{2} + k\pi\\\pi\sin x = k2\pi\\\pi\sin x = \pi + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\sin x = \dfrac{1}{2} + k\\\sin x = 2k\\\sin x = 1+ 2k\end{array}\right. \quad (k \in \Bbb Z)$
Do $- 1 \leq \sin x \leq 1$
nên $\left[\begin{array}{l}- 1 \leq \dfrac{1}{2} + k \leq 1 \\- 1 \leq 2k \leq 1\\-1\leq 1+ 2k \leq 1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}k = 0 \\k = 0\\k = -1\\k = 1\end{array}\right.$
Ta được:
$\left[\begin{array}{l}\sin x = \dfrac{1}{2} \\\sin x = 0\\\sin x = -1\\\sin x = 1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + n2\pi\\x = \dfrac{5\pi}{6} + n2\pi \\x = n\pi\\x = \dfrac{\pi}{2} + n\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + n2\pi\\x = \dfrac{5\pi}{6} + n2\pi\\x = n\dfrac{\pi}{2}\end{array}\right.\quad (n \in \Bbb Z)$
