решите уравнение sin^2x+sin^2 2x-sin^2 3x=sin^2 4x

Sin²x+sin²2x-sin²3x-sin²4x=0

1/2 (1-cos2x) + 1/2 (1-cos4x) - 1/2 (1-cos6x) - 1/2 (1-cos8x) = 0

1-cos2x+1-cos4x-1+cos6x-1+cos8x=0

(cos8x-cos2x) + 9cos6x-cos4x) = 0

-2sin3xsin5x-2sin2xsin5x=0

-2sin5x (sin3x+sin2x) = 0

-4sin5xsin2,5xcos0,5x=0

sin5x=0⇒5x=π4⇒x=πn/5

sin2,5x=0⇒2,5x=πn⇒x=2π/5

cos0.5x=0⇒0,5x=π/2+πn⇒x=π+2πn

ответ x=πn/5, x=π+2πn