Question

Помогите решить ур-е sinx + sin^2x/2= cos^2x/2и отобрать корни на промежутке[-2пи, -пи/2]

Answer 1

2sinx=cos²x-sin²x
sinx=cos2x
cos2x-cos(π/2-x)=0
-2sin(3x/2-π/4)sin(x/2+π/4)=0
sin(3x/2-π/4)=0 U sin(x/2+π/4)=0
3x/2-π/4=πn U x/2+π/4=πn
3x/2=π/4+πn u x/2=-π/4+πn
x=π/6+2πn/3 U x=-π/2+2πn
x=-11π/12;-7π/12;-π/2∈[-2π;-π/2]