a,
ĐK: $x\ne \dfrac{k\pi}{2}$
$8\cos x=\dfrac{\sqrt3}{\sin x}+\dfrac{1}{\cos x}$
$\to 8\sin x\cos x\cos x=\sqrt3\cos x+\sin x$
$\to 4\sin2x\cos x=2\sin\left(x+\dfrac{\pi}{3}\right)$
$\to 2\sin2x.\cos x=\sin\left(x+\dfrac{\pi}{3}\right)$
$\to \sin3x-\sin x=\sin\left(x+\dfrac{\pi}{3}\right)$
$\to \sin3x=\sin x+\sin\left(x+\dfrac{\pi}{3}\right)$
$\to \sin3x=2\sin\left(x+\dfrac{\pi}{6}\right).\cos\dfrac{\pi}{6}$
$\to \sin3x=\sin\left(x+\dfrac{\pi}{6}\right)$
$\to \left[ \begin{array}{l}3x=x+\dfrac{\pi}{6}+k2\pi \\ 3x=\pi-x-\dfrac{\pi}{6}+k2\pi \end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{12}+k\pi \\x=\dfrac{5\pi}{24}+\dfrac{k\pi}{2} \end{array} \right.$ (TM)
b,
$3\sin x-1=4\sin^3x+\sqrt3\cos x$
$\to 4\sin^3x-3\sin x+\sqrt3\cos3x=-1$
$\to -\sin3x+\sqrt3\cos3x=-1$
$\to \sin3x-\sqrt3\cos3x=1$
$\to 2\sin\left(3x-\dfrac{\pi}{3}\right)=1$
$\to \sin\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}$
$\to \left[ \begin{array}{l}3x-\dfrac{\pi}{3}= \dfrac{\pi}{6}+k2\pi \\ 3x-\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi \end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3} \\x=\dfrac{7\pi}{18}+\dfrac{k2\pi}{3} \end{array} \right.$
