Question

2 sin^2x+2sinx = sqrt(3)+ sqrt(3)sinx

Answer 1

$2\sin^2x+2\sin x= \sqrt{3} +\sqrt{3}\sin x \\ 2\sin^2 x+2\sin x-(\sqrt{3}+\sqrt{3}\sin x)=0 \\ 2\sin x(\sin x+1)-\sqrt{3}(1+\sin x)=0 \\ (\sin x+1)(2\sin x-\sqrt{3})=0 \\ \left[\begin{array}{ccc}\sin x=-1\\\sin x= \frac{\sqrt{3}}{2} \end{array}\right\to \left[\begin{array}{ccc}x_1=- \frac{\pi}{2} +2 \pi k,k \in Z\\ x_2=(-1)^k\cdot \frac{\pi}{3}+ \pi k,k \in Z \end{array}\right$