Question

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

Answer 1

$2\sin 2x\cos x+2\cos x=2\sin x\cos x+2{\cos }^{2}x$
Cancel $2\cos x$ and $\cos x=0$
gives $x=(n+\frac{1}{2})\pi ,n\in I$ .$\left(1\right)$
$\therefore \sin 2x+1=\sin x+\cos x$
or $(\sin x+\cos x{)}^2=\sin x+\cos x$
$\therefore \sin x+\cos x=0$ or $1$
$\sin x+\cos x=0$ gives $\tan x=-1=-\tan \frac{\pi }{4}$
$\therefore x=n\pi -\frac{\pi }{4}$ ..$\left(2\right)$
$\sin x+\cos x=1$
$\therefore x=2n\pi$ or $(2n\pi +\frac{\pi }{2})$ i.e., $(2n+\frac{1}{2})\pi$ $\left(3\right)$
But the second form is included in $\left(1\right)$. Hence the required answers are $(n+\frac{1}{2})\pi$, $n\pi -\frac{\pi }{4}$ and $2n\pi$ from $\left(1\right),\left(2\right),\left(3\right)$.