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алисенью
@алисенью
July 2022
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решите уравнение
2cos^2 x - 3sin x=0
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kirichekov
Verified answer
2cos²x-3sinx=0
2*(1-sin²x)-3sinx=0
2sin²x+3sinx-2=0
sinx=t, t∈[-1;1]
2t²+3t-2=0
D=3²-4*2*(-2)=25
t₁=(-3-5)/4, t₁=-2, -2∉[-1;1]
t₂=(-3+5)/4, t₂=1/2
sinx=1/2
x=(-1)^n*arcsin(1/2)+πn, n∈Z
x=(-1)^n*(π/6)+πn, n∈Z
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Answers & Comments
Verified answer
2cos²x-3sinx=02*(1-sin²x)-3sinx=0
2sin²x+3sinx-2=0
sinx=t, t∈[-1;1]
2t²+3t-2=0
D=3²-4*2*(-2)=25
t₁=(-3-5)/4, t₁=-2, -2∉[-1;1]
t₂=(-3+5)/4, t₂=1/2
sinx=1/2
x=(-1)^n*arcsin(1/2)+πn, n∈Z
x=(-1)^n*(π/6)+πn, n∈Z