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Sasha102
@Sasha102
July 2022
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помогите решить №2,3,4
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pomoshnik09
2)1-cjs²a/cos²a-cos2a=sin²a/cos²a-cos²a+sin²a=si²a/sin²a=1
sin2x-2cosx=0
2sinx*cosx-2cosx=0
2cosx(sinx-1)=0
cosx=0 x=π/2+πn. n∈z
sinx-1=0 sinx=1 x=π/2+2πk. k∈z
объединив серии корней x=π/2+πn. n∈z
4) cos2x=cos²x-sin²x=cos²x-(4/5)²=cos²x-16/25
найдем cos²x из основного тригонометрического тождества
сos²x+sin²x=1 cos²x=1-sin²x=1-16/25=9/25
cos 2x=9/25-16/25=-7/25
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Answers & Comments
sin2x-2cosx=0
2sinx*cosx-2cosx=0
2cosx(sinx-1)=0
cosx=0 x=π/2+πn. n∈z
sinx-1=0 sinx=1 x=π/2+2πk. k∈z
объединив серии корней x=π/2+πn. n∈z
4) cos2x=cos²x-sin²x=cos²x-(4/5)²=cos²x-16/25
найдем cos²x из основного тригонометрического тождества
сos²x+sin²x=1 cos²x=1-sin²x=1-16/25=9/25
cos 2x=9/25-16/25=-7/25