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