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kekw745
@kekw745
July 2022
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3cos^2x-sin^2x+4sinx=0
Решить уравнение. Если можно с объяснением)
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SolRu
3cos²x-sin²x+4sinx=0
cos²x=1-sin²x
3-3sin²x-sin²x+4sinx=0
-4sin²x+4sinx+3=0
t=sinx -1≤t≤1
-4t²+4t+3=0
D=16+48=64 √D=8
t₁=(-4+8)/(-8)= -(1/2)
t₂=(-4-8)/(-8)=12/8 не подходит т.к. (12/8)>1
sinx=-(1/2)
x=(-1)ⁿ(-π/6)+nπ, n∈Z
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Answers & Comments
cos²x=1-sin²x
3-3sin²x-sin²x+4sinx=0
-4sin²x+4sinx+3=0
t=sinx -1≤t≤1
-4t²+4t+3=0
D=16+48=64 √D=8
t₁=(-4+8)/(-8)= -(1/2)
t₂=(-4-8)/(-8)=12/8 не подходит т.к. (12/8)>1
sinx=-(1/2)
x=(-1)ⁿ(-π/6)+nπ, n∈Z