Решить уравнение: sin^2(2x)+sin^2(3x)+sin^2(4x)+sin^2(5x)=2

sin²2x+sin²3x +sin²4x +sin²5x =2 ; * * *cos2α =cos²α - sin²α = 1 -2sin²α ⇒2sin²α =1-cos2α ; sin²α =(1-cos2α)/2* * * * * * (1-cos4x)/2 +(1-cos6x)/2 +(1-cos8x)/2 +(1-cos10x) =2;* * *  * * *  удобно  сначала уравнение умножить на 2  * * * 2sin²2x+2sin²3x +2sin²4x +2sin²5x =2*2 ; (1-cos4x) +(1-cos6x) +(1-cos8x) +(1-cos10x) =4  ;  cos4x+cos6x +cos8x +cos10x  =0 ; cos6x+cos4x  +cos10x +cos8x =0 ; * * * cosα +cosβ =2cos(α+β)/2 * cos(α -β)/2  * * * 2cos5xcosx +2cos9x*cosx =0 ; 2cosx(cos9x+cos5x) =0 ; 4cosx*cos2x*cos7x =0  ; [cosx =0 ; cos2x =0; cos7x =0. cosx =0 ⇒ x =π/2+π*n ,n∈Z. cos2x =0 ⇒2x =π/2+π*n ⇔x =π/4+π/2*n ,n∈Z . cos7x =0  ⇒7x =π/2+π*n ⇔x =π/14+π/7*n ,n∈Z. ответ : π/2+π*n , π/4+π/2*n  , π/14+π/7*n , n∈Z.