Question

2-cos 2x+3sinx= 0
помогите,пожалуйста

Answer 1

$2-cos 2x+3sinx= 0 \2-(1-2sin^2x)+3sinx=0 \sinx=y, y in [-1;1] \2-1+2y^2+3y=0 \2y^2+3y+1=0 \D=9-8=1 \y_1= frac{-3+1}{4} =- frac{1}{2} \y_2= frac{-3-1}{4} =-1 \sinx=-1 \x_1=- frac{pi}{2}+2pi n, n in Z \sinx=- frac{1}{2} \ x_2= -frac{pi}{6} +2pi n, n in Z \x_3= -frac{5pi}{6} +2pi n, n in Z$