Решить уравнение sin(2x-7pi/2)+sin(3pi/2-8x)+cos(6x)=1

sin(2x -7π/2) +sin(3π/2 -8x) +cos6x =1;sin(-(7π/2 -2x)) -cos8x+cos6x =1 ;-sin(7π/2 -2x) -cos8x+cos6x =1  ;  * * *sin(7π/2 -2x) =  sin(4π-(π/2 +2x)) = -sin(π/2 +2x) = -cos2x * * *cos2x -cos8x +cos6x -1 =0 ;cos6x +cos2x -(1+cos8x) =0 ;  * * *или cos2x -cos8x  -(1-cos6x) =0  * * *2cos4xcos2x -2cos²4x =0 ;2cos4x(cos2x -cos4x) =0 ;2cos4x*2sinx*sin3x =0 ;4sinx*sin3x*cos4x=0 ;[sinx =0 ; sin3x =0 ; cos3x =0 .[ x=πk ; x=πk/3 ; 3x =π/2 +πk  , k∈Z.объединяя решении :[x = πk/3 ; x =π/6 + (π/3)* k   , k∈Z. ответ : πk/3 ;  x =π/6 +(π/3)* k , k∈Z.