То я туплю (cos3x+sin3x)^2=1+cos2x и второе sin^2(3x)+sin^2(81 пи - x)=1,5-sin^2(2x)

(cos3x +sin3x)² =1+cos2x ;
cos²3x +2cos3x*sin3x+sin²3x =1+cos2x ;
1 +sin6x =1+cos2x ;
cos(π/2 -6x) - cos2x =0 ;
cos(6x-π/2) - cos2x =0 ;
-2sin(2x -π/4)*sin(4x -π/4) =0 ; 
[sin(2x -π/4) =0 ; sin(4x -π/4)  =0 .⇒[ 2x -π/4 =πk ;4x -π/4=πk,k∈Z.

⇔[x =(π/8)(1 +4k) ; x =(π/16)(1+4k) , k∈Z.


sin²3x+sin²(81π - x)=1,5-sin²2x ; 
* * *sin(81π-x)=sin(40*2π+π-x) =sin(π-x)=sinx * * *
sin²3x+sin²x +sin²2x=1,5 ; 
(1-cos6x)/2+(1-cos2x)/2+(1-cos4x)/2=3/2 ; 
cos6x+cos2x+cos4x=0  ;
2cos4x*cos2x+cos4x=0 ;
2cos4x(cos2x+1/2)=0 ⇔[ cos4x =0 ; cos2x = -1/2 . 
[4x =π/2 +πk ,2x =± (π - π/3) +2πk , k∈Z.
[x =π/8 +(π/4)*k ,x =± π/3 +πk , k∈Z.