Mamy

$\left(4{\sin }^{2}x-1\right)\cdot \sin x={\cos }^{2}x-3{\sin }^{2}x$  

$\left(3{\sin }^{2}x+{\sin }^{2}x-1\right)\cdot \sin x={\cos }^{2}x-3{\sin }^{2}x$  

$\left(3{\sin }^{2}x-{\cos }^{2}x\right)\cdot \sin x={\cos }^{2}x-3{\sin }^{2}x$  

$\left(3{\sin }^{2}x-{\cos }^{2}x\right)\cdot \sin x-{\cos }^{2}x+3{\sin }^{2}x=0$  

$\left(3{\sin }^{2}x-{\cos }^{2}x\right)\left(\sin x+1\right)=0$  

$3{\sin }^{2}x-{\cos }^{2}x=0\text{lub}\sin x+1=0$  

$3{\sin }^{2}x-\left(1-{\cos }^{2}x\right)=0\text{lub}\sin x+1=0$  

$3{\sin }^{2}x-1+{\sin }^{2}x=0\text{lub}\sin x+1=0$  

$4{\sin }^{2}x-1=0\text{lub}\sin x+1=0$  

$\left(2\sin x-1\right)\left(2\sin x+1\right)=0\text{lub}\sin x+1=0$  

$2\sin x-1=0\text{lub}2\sin x+1=0\text{lub}\sin x+1=0$  

$\sin x=\frac{1}{2}\text{lub}\sin x=-\frac{1}{2}\text{lub}\sin x=-1$  

$\sin x=\frac{1}{2}\text{lub}\sin \left(-x\right)=\frac{1}{2}\text{lub}\sin x=-1$  

$x=\frac{\pi }{6}+2k\pi \text{lub}x=\frac{5\pi }{6}+2k\pi ,k\in \mathbb{Z}$  

lub

$x=-\frac{\pi }{6}+2k\pi \text{lub}x=-\frac{5\pi }{6}+2k\pi ,k\in \mathbb{Z}$  

lub

$x=\frac{3\pi }{2}+2k\pi ,k\in \mathbb{Z}$  

W przedziale mamy

$x=\left\{-\frac{\pi }{2},-\frac{5\pi }{6},-\frac{\pi }{6}\right\}$