Question

Solve: ${\cos }^{2}x-2\cos x=4\sin x-\sin 2x,0\le x\le \pi$.

Answer 1

$\cos x(\cos x-2)+2\sin x(\cos x-2)=0$

$(\cos x+2\sin x)(\cos x-2)=0$ but $\cos x\ne 2$

$\therefore \tan x=-\frac{1}{2}=\tan \alpha$

$\therefore x=n\pi +\alpha$

$x=n\pi +{\tan }^{-1}(-\frac{1}{2})$

$\therefore$ In the given interval $0\le x\le \pi$

$\pi +{\tan }^{-1}(-\frac{1}{2})=\pi -{\tan }^{-1}\left(\frac{1}{2}\right)$.