Sin^2(x)+sin^2(2x)+sin^2(3x)+sin^2(4x)=2 cos^3(x)+sin^4(x)=1 ПОМОГИТЕ РЕШИТЬ ОЧЕНЬ СРОЧНО. Created by Vetta1605. algebra-ru

Sin^2(x)+sin^2(2x)+sin^2(3x)+sin^2(4x)=2 cos^3(x)+sin^4(x)=1 ПОМОГИТЕ РЕШИТЬ ОЧЕ...

Vetta1605 @Vetta1605

July 2022 1 4 Report

Sin^2(x)+sin^2(2x)+sin^2(3x)+sin^2(4x)=2

cos^3(x)+sin^4(x)=1

ПОМОГИТЕ РЕШИТЬ ОЧЕНЬ СРОЧНО

Answers & Comments

sedinalana

Verified answer

1
sin²2x+sin²4x=(1-sin²x)+(1-sin²3x)
sin²2x+sin²4x=cos²x+cos²3x
(1-cos4x)/2+(1-cos8x)/2=(1+cos2x)/2+(1+cos6x)/2
1-cos4x+1-cos8x=1+cos2x+1+cos6x
cos4x+cos8x+cos2x+cos6x=0
2cos6xcos2x+2cos4xcos2x=0
2cos2x(cos6x+cos4x)=0
2cos2x*2cosx*cos5x=0
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2,n∈z
cosx=0⇒x=π/2+πn,n∈z
cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5
2
cos³x+(sin^4x-1)=0
cos³x+(sin³x-1)(sin²x+1)=0
cos³x-cos²x(1+sin²x)=0
cos²x(cosx-1-sin²x)=0
cos²x(cosx-1-1+cos²x)=0
cos²x(cos²x+cosx-2)=0
cosx=0πx=π/2+πn,n∈z
cos²x+cosx-2=0
cosx=a
a²+a-2=0
a1+a2=-1 U a1*a2=-2
a1=-2⇒cosx=-2<-1 нет решения
a2=1⇒cosx=1⇒x=2⇒n,n∈z

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