1)
$2\sin^2x+\sin x-1=0$
$\to \left[ \begin{array}{l}\sin x=-1 \\ \sin x=\dfrac{1}{2}\end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{-\pi}{2}+k2\pi \\x=\dfrac{\pi}{6}+k2\pi\\ x=\dfrac{5\pi}{6}+k2\pi \end{array} \right.$
2)
$3\cos2x-2\cos x-8=0$
$\to 3(2\cos^2x-1)-2\cos x-8=0$
$\to 6\cos^2x-2\cos x-11=0$
$\to \cos x=\dfrac{1\pm\sqrt{67}}{6}\notin[-1;1]$ (loại)
Vậy PT vô nghiệm
3)
$\sqrt3\sin x+\cos x=1$
$\to 2\sin\left(x+\dfrac{\pi}{6}\right)=1$
$\to \sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{1}{2}$
$\to \left[ \begin{array}{l}x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi \\x+\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi \end{array} \right.$
$\to \left[ \begin{array}{l}x=k2\pi \\x=\dfrac{2\pi}{3}+k2\pi \end{array} \right.$
4)
$\sin4x-\sin6x=\sqrt3(\cos6x+\cos4x)$
$\to \sin4x-\sqrt3\cos4x=\sin6x+\sqrt3\cos6x$
$\to \dfrac{1}{2}\sin4x-\dfrac{\sqrt3}{2}\cos4x=\dfrac{1}{2}\sin6x+\dfrac{\sqrt3}{2}\cos6x$
$\to \sin\left(4x-\dfrac{\pi}{3}\right)=\sin\left( 6x+\dfrac{\pi}{3}\right)$
$\to \left[ \begin{array}{l}6x+\dfrac{\pi}{3}=4x-\dfrac{\pi}{3}+k2\pi \\6x+\dfrac{\pi}{3}=\pi-4x+\dfrac{\pi}{3}+k2\pi \end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{-\pi}{3}+k\pi \\x=\dfrac{\pi}{10}+\dfrac{k\pi}{5} \end{array} \right.$
