Sin²x - cos²x = cos4x ⇔ - (cos²x - sin²x) = cos4x ⇔ -cos2x =cos4x ⇔
0 =cos4x+cos2x ⇔2cos(4x -2x)/2 *cos(4x+2x) /2 =0 ⇔2cosx*cos3x =0 ⇒
[ cosx =0 ; cos3x =0 . ⇒ соs, 3x =π/2 +πn ,n∈Z. ⇔
[ x =π/2 +πk , x =π/6 +(π/3)*n ,k, n∈Z. ⇒ x =π/6 +(π/3)*n , n∈Z.
серия решения x = π/6 +(π/3)*n , n∈Z содержит и решения π/2 +πk , k∈Z при n =1 +3k .
ответ : π/6 +(π/3)*n , n∈Z.
* * * * * * *
cosx =0 ⇒ x = π/2 +πk , k∈Z
cos3x =0 ⇒ 3x = π/2 +πn ,n ∈Z ⇔ x = π/6 +πn /3 ,n ∈Z .
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π/2 +πk = π/6 +πn /3 ⇔3 +6k =1 +2n ⇔ n =1 +3k
--- или по другому:
cos3x =cosx(4cos²x -3)