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Acaciadealbat
@Acaciadealbat
July 2022
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2sin2x-sin^2x=3cos^2x
Найдите решение тригонометрического неравенства
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DariosI
Verified answer
2sin2x-sin^2x-3cos^2x=0
4sinx*cosx-sin²x-3cos²x=0 :cos²x
4sinx*cosx/cos²x-sin²x/cos²x-3cos²x/cos²x=0
4sinx/cosx-
sin²x/cos²x-3=0
4tgx-tg
²x-3=0
tgx=y
4y-y²-3=0
y²-4y+3=0
D=16-4*3=4
y₁=(4-2)/2=1
y₂=(4+2)/2=3
tgx=1
x=(-1)ⁿπ/4+πn, n∈Z
tgx=3
x=
(-1)ⁿarctgx+πn, n∈Z
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Acaciadealbat
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Answers & Comments
Verified answer
2sin2x-sin^2x-3cos^2x=04sinx*cosx-sin²x-3cos²x=0 :cos²x
4sinx*cosx/cos²x-sin²x/cos²x-3cos²x/cos²x=0
4sinx/cosx-sin²x/cos²x-3=0
4tgx-tg²x-3=0
tgx=y
4y-y²-3=0
y²-4y+3=0
D=16-4*3=4
y₁=(4-2)/2=1
y₂=(4+2)/2=3
tgx=1
x=(-1)ⁿπ/4+πn, n∈Z
tgx=3
x=(-1)ⁿarctgx+πn, n∈Z