Question

$(2\sin x-\cos x)(1+\cos x)={\sin }^{2}x$
If true then enter $1$ and if false then enter $0$

Answer 1

Given, $(2\sin x-\cos x)(1+\cos x)={\sin }^{2}x$
$\Rightarrow (2\sin x-\cos x)(1+\cos x)-{\sin }^{2}x=0$
$\Rightarrow (2\sin x-\cos x)(1+\cos x)-(1-\cos x)(1+\cos x)=0$
$\Rightarrow (1+\cos x)(2\sin x-1)=0$
$\Rightarrow 1+\cos x=0or2\sin x-1=0$
$\Rightarrow \cos x=-1or\sin x=\frac{1}{2}\Rightarrow x=(2n+1)\pi ,n\in Ior\sin x=\sin \frac{\pi }{6}$
$\Rightarrow x=n\pi +{(-1)}^n\frac{\pi }{6},n\in I$
$\therefore$ Solution of given equation is $(2n+1)\pi ,m\in Iorn\pi +{(-1)}^n\frac{\pi }{6},n\in I$