Помогите решить тригонометрическое уравнение! cos4x + 2sin^2x = ...

Ira

cos(4x)+2*(sin(x))^2=01-2*(sin(2x))^2+2*(sin(x))^2=01-8*(sin(x))^2*(cos(x))^2+2*(sin(x))^2=01-8*(sin(x))^2*(1-sin(x))^2)+2*(sin(x))^2=01-8y(1-y)+2y=08y^2-6y+1=0y1=1/2 y2=1/4(sin(x))^2=1/2sin(x)=sqrt(2)/2 x1=(-1)^k*п/4+пк кЄZsin(x)=-sqrt(2)/2 x2=(-1)^(m+1)*п/4+пm mЄZ(sin(x))^2=1/4sin(x)=1/2 x3=(-1)^l*п/6+пl lЄZsin(x)=-1/2 x4=(-1)^(s+1)*п/6+пs sЄZ