Решите уравнение!

4sin^2x (1+cos2x) = 1-cos2x

4 * (1-cos2x) / 2 * (1+cos2x) = 1-cos2x

2 * (1-cos2x) * (1+cos2x) - (1-cos2x) = 0

(1-cos2x) (2+2cos2x-1) = 0

(1-cos2x) (2cos2x+1) = 0

1-cos2x=0⇒cos2x=1⇒2x=2πn⇒x=πn

cos2x=-1/2⇒2x=+-2π/3+2πn⇒x=+-π/3+πn