Помогите!!!!! sin^2x-cos^2x=cos4x. Created by irdinkey. algebra-ru

Помогите!!!!! sin^2x-cos^2x=cos4x...

irdinkey @irdinkey

June 2022 1 16 Report

Помогите!!!!!
sin^2x-cos^2x=cos4x

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SiMpOCh

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Sin²x-cos²x=cos(4x)←←←
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 равен:ответ : π/6 +(π/3)*n.
Наверно так

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