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servicepack21
@servicepack21
August 2022
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Решите одно уравнение, только очень-очень подробно
3sin2x+cos2x=cos^2x
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kirichekov
Verified answer
3sin2x+cos2x=cos²x
3*(2sinx*cosx)+(cos²x-sin²x)=cos²x
6sinx*cosx+cos²x-sin²x-cos²x=0
6sinx*cosx-sinx²x=0 |:cos²x≠0
(6sinx*cos)/cos²x-sin²x/cos²x=0
6tgx-tg²x=0. tgx*(6-tgx)=0
tgx=0 или 6-tgx=0
1. tgx=0,
x
₁
=πn, n∈Z
2. 6-tgx=0, tgx=6.
x₂=arctg6+πn, n∈Z
2 votes
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Answers & Comments
Verified answer
3sin2x+cos2x=cos²x3*(2sinx*cosx)+(cos²x-sin²x)=cos²x
6sinx*cosx+cos²x-sin²x-cos²x=0
6sinx*cosx-sinx²x=0 |:cos²x≠0
(6sinx*cos)/cos²x-sin²x/cos²x=0
6tgx-tg²x=0. tgx*(6-tgx)=0
tgx=0 или 6-tgx=0
1. tgx=0, x₁=πn, n∈Z
2. 6-tgx=0, tgx=6. x₂=arctg6+πn, n∈Z