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VOLKANCHIK
@VOLKANCHIK
June 2022
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Сегодня срочно!!!! 1+2cosx=sin2x+2sinx x€[-7/2П;-2П]
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sharadi
Verified answer
1+2cosx=sin2x+2sinx x€[-7/2
π
;-2
π
]
1 + 2Cosx - Sin2x - 2Sinx = 0
(1 - Sin2x) + ( 2Cosx -2Sinx) = 0
(Sin
² x + Cos²x - 2SinxCosx) + 2(Cosx -Sinx) = 0
(Cosx -Sinx)² + 2(Cosx -Sinx) = 0
(Cosx - Sinx)(Cosx -Sinx +2) = 0
Cosx - Sinx = 0 |:Cosx или Cosx -Sinx +2 = 0
1 - tgx = 0 Cosx -Sinx = -2
tgx = 1 ∅
x = π/4 + πk , k ∈ Z
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Answers & Comments
Verified answer
1+2cosx=sin2x+2sinx x€[-7/2π;-2π]1 + 2Cosx - Sin2x - 2Sinx = 0
(1 - Sin2x) + ( 2Cosx -2Sinx) = 0
(Sin² x + Cos²x - 2SinxCosx) + 2(Cosx -Sinx) = 0
(Cosx -Sinx)² + 2(Cosx -Sinx) = 0
(Cosx - Sinx)(Cosx -Sinx +2) = 0
Cosx - Sinx = 0 |:Cosx или Cosx -Sinx +2 = 0
1 - tgx = 0 Cosx -Sinx = -2
tgx = 1 ∅
x = π/4 + πk , k ∈ Z