Question

Solve the following equation:
$3{\cos }^{2}x-{\sin }^{2}x-\sin 2x=0.$

Answer 1

${3\cos }^{2}x-2\sin x\cos x-{\sin }^{2}x=0$

${3\cos }^{2}x-3\sin x\cos x+\sin x\cos x-{\sin }^{2}x=0$

$(3\cos x+\sin x)(\cos -\sin x)=0$

So, $\tan x=1$ and $\tan x=-3.$

$\therefore$ $x=n\pi +\frac{\pi }{4}$ and $x=n\pi -{\tan }^{-1}(-3).$