Решите cos2x +10sinx -9=0

[latex]cos2x+10sinx-9=0 \\ cos^{2}x-sin^{2}x+10sinx-9=0 \\ (1-sin^{2}x)-sin^{2}x+10sinx-9=0 \\ -2sin^{2}x+10sinx-8=0 \\ sinx=t \\ -2t^{2}+10t-8=0 |:(-2) \\ t^{2}-5t+4=0 \\ D=(-5)^{2}-4*4=25-16=9 \\ \\ x_{1}= \frac{5+3}{2}=4 \\ x_{2}= \frac{5-3}{2}=1 \\ \\ sinx \neq 4 \\ sinx=1 \\ x= \frac{ \pi }{2}+2 \pi n[/latex] Ответ: x=п/2+2пn