Ответ:
Объяснение:
2cos2x=2(cos^2x+sin^2x)(cos^2x-sin^2x)=2(cos^4x-sin^4x)
sin^4x+cos^4x=2cos^4x-2sin^4x
3sin^4x=cos^4x
tg^4x=1/3
tgx=(1/3)^1/4
x=arctg((1/3)^1/4)
sin^2x/cos^2x=1/√3
sin^2x/(1-sin^2x)=1/√3
√3sin^2x=1-sin^2x
sin^2x=1/(√3+1)=(√3-1)/2
cos2x=1-2sin^2x=1-√3+1=2-√3
2x=+-arccos(2-√3)+2Пk
x=+-1/2arccos(2-√3)+Пk