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miliaushaahyar
@miliaushaahyar
July 2022
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ПОМАГИТЕ ПЛИЗЗЗ решите уравнение 4sinx+3cos^2x=sin^2x
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m11m
Verified answer
4sinx+3(1-sin²x)-sin²x=0
4sinx+3-3sin²x-sin²x=0
-4sin²x+4sinx+3=0
4sin²x-4sinx-3=0
t=sinx
t²=sin²x
4t²-4t-3=0
D=(-4)²-4*4*(-3)=16+48=64=8²
t₁=(4-8)/8=-4/8= -1/2
t₂=(4+8)/8=12/8=1.5
При t= -1/2
sinx= -1/2
x=(-1)ⁿ⁺¹ п/6 + пn, n∈Z
При t=1.5
sinx=1.5
1.5>1
нет решений
Ответ: (-1)ⁿ⁺¹ п/6 + пn, n∈Z.
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Answers & Comments
Verified answer
4sinx+3(1-sin²x)-sin²x=04sinx+3-3sin²x-sin²x=0
-4sin²x+4sinx+3=0
4sin²x-4sinx-3=0
t=sinx
t²=sin²x
4t²-4t-3=0
D=(-4)²-4*4*(-3)=16+48=64=8²
t₁=(4-8)/8=-4/8= -1/2
t₂=(4+8)/8=12/8=1.5
При t= -1/2
sinx= -1/2
x=(-1)ⁿ⁺¹ п/6 + пn, n∈Z
При t=1.5
sinx=1.5
1.5>1
нет решений
Ответ: (-1)ⁿ⁺¹ п/6 + пn, n∈Z.