Question

Sin2x+2cosx=sinx+1 cos2x+sinx=cos^2x

Answer 1

$\sin2x+2\cos x=\sin x+1\\ 2\sin x\cos x+2\cos x=\sin x+1\\ 2\cos x(\sin x+1)=\sin x+1\\ (\sin x+1)(2\cos x-1)=0\\ \left[\begin{array}{ccc}\sin x=-1\\ \cos x=0.5\end{array}\right\Rightarrow~~~~ \left[\begin{array}{ccc}x_1=- \frac{\pi}{2}+2 \pi k,k \in \mathbb{Z}\\ x_2=\pm \frac{\pi}{3}+2 \pi n,n \in \mathbb{Z} \end{array}\right$

$\cos 2x+\sin x=\cos^2x\\ 1-2\sin^2x+\sin x=1-\sin^2x\\ \sin^2x-\sin x=0\\ \sin x(\sin x-1)=0\\ \left[\begin{array}{ccc}\sin x=0\\ \sin x=1\end{array}\right~~\Rightarrow~~~~~ \left[\begin{array}{ccc}x_1= \pi k,k \in \mathbb{Z}\\ x_2= \frac{\pi}{2}+2 \pi k,k \in \mathbb{Z} \end{array}\right$