Đáp án: $$\left[\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.$$
Giải thích các bước giải:
$\cos6x.\cos2x=\sin7x.\sin3x$
$\Leftrightarrow \dfrac{1}{2}\Big( \cos8x+\cos4x\Big)=\dfrac{-1}{2}\Big(\cos10x-\cos4x\Big)$
$\Leftrightarrow \cos8x+\cos4x=-\cos10x+\cos4x$
$\Leftrightarrow \cos10x+\cos8x=0$
$\Leftrightarrow 2\cos9x.\cos x=0$
$\Leftrightarrow $ $$\left[\begin{matrix}\cos9x=0\\\cos x=0\end{matrix}\right.$$
$\Leftrightarrow$ $$\left[\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.$$
