То я туплю (cos3x+sin3x)^2=1+cos2x и второе sin^2(3x)+sin^2(81 пи - x)=1,5-sin^2(2x)
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Ответы на вопрос:
(cos3x +sin3x)² =1+cos2x ; cos²3x +2cos3x*sin3x+sin² 3x =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.
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