2cos^2x=11sinx+7 пожалуйстаа решите

[latex]2cos ^{2} x-11sinx-7=0\\2-2sin ^{2} x-11sinx-7=0\\-2sin ^{2} x-11sinx-5=0\\2sin ^{2} x+11sinx+5=0\\sinx=t\\-1 \leq t \leq 1 \\ 2t ^{2} +11t+5=0\\D=121-40=81\\ \sqrt{D} =9\\ t_{1} = \frac{-11+9}{4} =- \frac{1}{2} \\ t _{2} = \frac{-11-9}{4} =-5 \\ [/latex] [latex]\\ sinx=- \frac{1}{2} \\ \\ x=(-1) ^{n+1} \frac{ \pi }{6} + \pi n[/latex] n ∈ Z