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luffi
@luffi
July 2022
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Найти количество корней уравнения 2tg^2x+3=3\cosx , принадлежащих отрезку [0;360]
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Найти количество корней уравнения 2tg^2x+3=3\cosx , принадлежащих отрезку [0;360°
]
одз :
cosx ≠0 x≠π/2+πn, n∈Z.
2tg²x+3=3\cosx 2sin²x/cos²x=3cosx/cos²x ⇔ 2sin²x=3cosx⇔
2(1-cos²x)=3cosx 2cos²x+3cosx -2=0
cosx =t ItI≤1, t≠0 2t²+3t-2=0
t1=[-3-√(9+16)] /2 =-4 посторонний корень,
t2=[-3+√(9+16)] /2 =1
cosx =1 ⇔x=2πn, n∈Z,
x∈[0;360°] : x=0°, x=360°
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Verified answer
Найти количество корней уравнения 2tg^2x+3=3\cosx , принадлежащих отрезку [0;360°]одз : cosx ≠0 x≠π/2+πn, n∈Z.
2tg²x+3=3\cosx 2sin²x/cos²x=3cosx/cos²x ⇔ 2sin²x=3cosx⇔
2(1-cos²x)=3cosx 2cos²x+3cosx -2=0
cosx =t ItI≤1, t≠0 2t²+3t-2=0
t1=[-3-√(9+16)] /2 =-4 посторонний корень,
t2=[-3+√(9+16)] /2 =1
cosx =1 ⇔x=2πn, n∈Z,
x∈[0;360°] : x=0°, x=360°