Помогите! 4-cos^2*2x=3sin^2*2x+2sin4x. Created by 12345sergey12345. algebra-ru

Помогите! 4-cos^2*2x=3sin^2*2x+2sin4x...

July 2022 1 7 Report

Помогите!
4-cos^2*2x=3sin^2*2x+2sin4x

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kovalevdmitriy0 2*(1+сos4x)/2 *3sin4x+4*(1-cos4x)/2=0
1+cos4x+3sin4x+2-2cos4x=0
1-cos4x+3sin4x=0
2sin²2x+6sin2xcos2x=0/2cos²2x≠0
tg²2x+3tg2x=0
tg2x(tg2x+3)=0
tg2x=0⇒2x=πn⇒x=πn/2
tg2x=-3⇒2x=-arctg3+πn⇒x=-1/2arctg3+πn/2

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12345sergey12345 August 2021 | 0 Ответы

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12345sergey12345 July 2021 | 0 Ответы

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