a) $\sin6x + \sin5x + \sin x = 0$
$\Leftrightarrow \sin6x + 2\sin3x\cos2x = 0$
$\Leftrightarrow 2\sin3x\cos3x + 2\sin3x\cos2x = 0$
$\Leftrightarrow \sin3x(\cos3x + \cos2x) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin3x = 0\\\cos3x = \cos(\pi - 2x)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}3x = k\pi\\3x = \pi - 2x + k2\pi\\3x = 2x - \pi + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{3}\\x = \dfrac{\pi}{5} + k\dfrac{2\pi}{5}\\x =- \pi + k2\pi\end{array}\right. \quad (k \in \Bbb Z)$
b) $\sin6x + \cos8x - \cos4x = 0$
$\Leftrightarrow \sin6x - 2\sin6x\sin2x = 0$
$\Leftrightarrow \sin6x(1 - 2\sin2x) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin6x = 0\\\sin2x = \dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}6x = k\pi\\2x = \dfrac{\pi}{6}+ k2\pi\\2x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{6}\\x = \dfrac{\pi}{12}+ k\pi\\x = \dfrac{5\pi}{12} + k\pi\end{array}\right. \quad (k \in \Bbb Z)$
c) $(1 + \cos x)(2\sin x - \cos x) = \sin^2x$
$\Leftrightarrow (1 + \cos x)(2\sin x - \cos x) = 1 - \cos^2x$
$\Leftrightarrow (1 + \cos x)(2\sin x - \cos x) = (1 - \cos x)(1 + \cos x)$
$\Leftrightarrow (1 + \cos x)(2\sin x - \cos x - 1 + \cos x) = 0$
$\Leftrightarrow (1 + \cos x)(2\sin x - 1) = 0$
$\Leftrightarrow \left[\begin{array}{l}\cos x = -1\\\sin x = \dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\x = \dfrac{\pi}{6}+ k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right. (k \in \Bbb Z)$
d) $tan^2x = \dfrac{1 + \cos x}{1 - \sin x}\qquad (*)$
$ĐK: \, \begin{cases}\cos x \ne 0\\\sin x \ne 1\end{cases} \Leftrightarrow x \ne \dfrac{\pi}{2} + k\pi$
$(*) \Leftrightarrow \dfrac{1 - \cos^2x}{1 - \sin^2x} - \dfrac{1 +\cos x}{1 - \sin x} = 0$
$\Leftrightarrow \dfrac{(1 - \cos x)(1 + \cos x)}{1 + \sin x} - (1 + \cos x) = 0$
$\Leftrightarrow (1 + \cos x)\left(\dfrac{1 - \cos x}{1 + \sin x} -1\right)= 0$
$\Leftrightarrow \left[\begin{array}{l}\cos x = -1\\1 - \cos x = 1 + \sin x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\cos x = -1\\\sin x + \cos x = 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\\sin\left(x + \dfrac{\pi}{4}\right) = 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \pi + k2\pi\\x =- \dfrac{\pi}{4}+ k\pi\end{array}\right. \quad (k \in \Bbb Z)$
e) $4\cos^2x - 2\cos2x - 2\cos^22x = 0$
$\Leftrightarrow 2(1+ \cos2x) - 2\cos2x - 2\cos^22x = 0$
$\Leftrightarrow \cos^22x -1 = 0$
$\Leftrightarrow 1 + \cos4x = 2$
$\Leftrightarrow \cos4x = 1$
$\Leftrightarrow 4x = k2\pi$
$\Leftrightarrow x = k\dfrac{\pi}{2} \quad (k \in \Bbb Z)$
