Question

Решите cos2x +10sinx -9=0

Answer 1

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

Ответ: x=п/2+2пn