1/sin^2x+ 3/cos(15П/2+x)=-2

$\frac{1}{\sin^2x}+ \frac{3}{\cos(\frac{15\pi}{2}+x)}=-2\\\frac{1}{\sin^2x}+ \frac{3}{\sin x}=-2\mid \cdot \sin^2 x\ (\sin x\ne 0)\\1+3\sin x=-2\sin^2x\\2\sin^2x+3\sin x+1=0\\t=\sin x,\ |t| \leq1\\2t^2+3t+1=0\\D=9-8=1\\\left[ \begin{array}{ccc} t_1=\frac{-3+1}{4}=-\frac{1}{2} \\ t_2=\frac{-3-1}{4}=-1\\ \end{array} \left[ \begin{array}{ccc} \sin x=-\frac{1}{2} \\ \sin x=-1\\ \end{array}\$
$\left[ \begin{array}{ccc} x_1=-\frac{\pi}{6}+2\pi n \\ x_2=\frac{5\pi}{6}+2\pi n\\ x_3=-\frac{\pi}{2}+2\pi n \end{array} \left[ \begin{array}{ccc} x_1=-\frac{\pi}{6}+\pi n\\ x_2=-\frac{\pi}{2}+2\pi n \end{array}$

Ответ: $\begin{cases} \left[ \begin{array}{ccc} x_1=-\frac{\pi}{6}+\pi n\\ x_2=-\frac{\pi}{2}+2\pi n \end{array} \\ x\ne \pi n \end{cases}$