Помогите решить уравнение!

sin^2 x + sin^2 2x + sin^2 3x + sin^2 4x = 2

умоляю! кто нибудь

(1-cos2x) / 2 + (1-cos4x) / 2 + (1-cos6x) / 2 + (1-cos8x) / 2=2

1-cos2x+1-cos4x+1-cos6x+1-cos8x=4

(cos2x+cos8x) + (cos4x+cos8x) = 0

2cos5xcos3x+2cos5xcos2x=0

2cos5x (cos3x+cos2x) = 0

2cos5x*2cos (5x/2) cos (x/2) = 0

cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5, n∈z

cos (3x/2) = 0⇒3x/2=π/2+πk⇒x=π/3+2πk/3, k∈z

cos (x/2) = 0⇒x/2=π/2+πm⇒x=π+2πm, m∈z