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servicepack21
@servicepack21
August 2022
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Подробное решение
4cos^2x-8 sin x+1=0
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kirichekov
Verified answer
4cos²x-8sinx+1=0
4*(1-sin²x)-8sinx+1=0
4sin²x+8sinx-5=0
замена переменных: sinx=t, t∈[-1;1]
4t²+8t-5=0. D=144. t₁=-2,5. -2,5∉[-1;1] посторонний корень
t₂=1/2
обратная замена:
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
4cos²x-8sinx+1=04*(1-sin²x)-8sinx+1=0
4sin²x+8sinx-5=0
замена переменных: sinx=t, t∈[-1;1]
4t²+8t-5=0. D=144. t₁=-2,5. -2,5∉[-1;1] посторонний корень
t₂=1/2
обратная замена:
sinx=1/2. x=(-1)^n *arcsin(1/2)+πn, n∈Z
x=(-1)^n*(π/6)+πn, n∈Z