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Liffkoil36
@Liffkoil36
June 2022
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2cos^4x=3 sin^2х-2 помогите кто может,даю 20 БАЛЛОВ!!!!
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DariosI
Verified answer
2cos⁴x=3sin²x-3+1
2cos⁴x=3(sin²x-cos²x-sin²x)+1
2cos⁴x=3(-cos²x)
+1
2cos⁴x+3cos²x-1=0
cos²x=y
2y²+3y-1=0
D=9+4*2=17
y=(3-√17)/4
y=(3+√17)/4 не подходит
cos²x=
(3-√17)/4
cosx=+-√(3-√17)/2
x=+-(arccos(
-√(3-√17)/2))+2
πn
x=+-(arccos(
√(3-√17)/2))+2
πn
1 votes
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Answers & Comments
Verified answer
2cos⁴x=3sin²x-3+12cos⁴x=3(sin²x-cos²x-sin²x)+1
2cos⁴x=3(-cos²x)+1
2cos⁴x+3cos²x-1=0
cos²x=y
2y²+3y-1=0
D=9+4*2=17
y=(3-√17)/4
y=(3+√17)/4 не подходит
cos²x=(3-√17)/4
cosx=+-√(3-√17)/2
x=+-(arccos(-√(3-√17)/2))+2πn
x=+-(arccos(√(3-√17)/2))+2πn