Sin^2 x-5ctgx=1+2sin2x

[latex]\sin^2x-5ctgx=1+2\sin2x \\ \sin^2x-5ctgx=\sin^2x+\cos^2x+2\sin2x \\ -5\cos x=\cos^2x\cdot \sin x+4\sin^2x\cos x \\ \cos x(\cos x\sin x+9\sin^2 x+5\cos^2x)=0 \\ \cos x=0 \\ x= \frac{\pi}{2} +\pi n,n \in Z[/latex] [latex]9\sin^2x+\sin x\cos x+5\cos^2x =0|:\cos^2 x \\ 9tg^2x+tg x+5=0 \\ D=1-9*5*4<0[/latex] Ответ: [latex]\frac{\pi}{2} +\pi n[/latex]