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kiskristina
@kiskristina
August 2022
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помогите решить уравнение cos4x=6cos^2x-5
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m11m
Verified answer
Cos4x=6cos²x - 5
cos(2*2x)=6cos²x -5
cos² 2x - sin² 2x=6cos²x - 5
cos² 2x - (1-cos² 2x)=6cos²x - 5
cos² 2x - 1 + cos² 2x=6cos²x - 5
2cos² 2x - 1=6cos²x - 5
2(cos²x - sin²x)² - 1=6cos²x - 5
2(2cos²x -1)² - 1=6cos²x - 5
2(4cos⁴x - 4cos²x+1)-1=6cos²x - 5
8cos⁴x - 8cos²x + 2-1 - 6cos²x +5 =0
8cos⁴x - 14cos²x +6=0
4cos⁴x - 7cos²x +3=0
y=cos²x
4y² -7y +3=0
D=49-48=1
y₁=(7-1)/8=6/8=3/4
y₂=(7+1)/8=1
При у=3/4
cos²x=3/4
a) cosx=√3/2
x=(+/-)
,
k∈Z
б) cosx=-√3/2
k∈Z
При у= 1
cos²x=1
a) cosx=1
x=2πk, k∈Z
б) cosx= -1
x=π+2πk, k∈Z
Ответ:
2πk;
π+2πk, k∈Z
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Answers & Comments
Verified answer
Cos4x=6cos²x - 5cos(2*2x)=6cos²x -5
cos² 2x - sin² 2x=6cos²x - 5
cos² 2x - (1-cos² 2x)=6cos²x - 5
cos² 2x - 1 + cos² 2x=6cos²x - 5
2cos² 2x - 1=6cos²x - 5
2(cos²x - sin²x)² - 1=6cos²x - 5
2(2cos²x -1)² - 1=6cos²x - 5
2(4cos⁴x - 4cos²x+1)-1=6cos²x - 5
8cos⁴x - 8cos²x + 2-1 - 6cos²x +5 =0
8cos⁴x - 14cos²x +6=0
4cos⁴x - 7cos²x +3=0
y=cos²x
4y² -7y +3=0
D=49-48=1
y₁=(7-1)/8=6/8=3/4
y₂=(7+1)/8=1
При у=3/4
cos²x=3/4
a) cosx=√3/2
x=(+/-),
k∈Z
б) cosx=-√3/2
k∈Z
При у= 1
cos²x=1
a) cosx=1
x=2πk, k∈Z
б) cosx= -1
x=π+2πk, k∈Z
Ответ:
2πk;
π+2πk, k∈Z