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swallow2303
@swallow2303
August 2022
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помогите пожалуйста
3sin2x+4cos2x=5
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1Catherine
3sin2x+4cos2x=5 |:5
3/5 sin2x + 4/5 cos 2x = 1
3/5=sinA, 4/5=cosA
sinA*sin 2x + cosA*cos 2x = 1
cos(2x - A) = 1
2x - A = +-arccos 1/2 + 2πn,n∈Z
2x - A = +-π/3 + 2πn,n∈Z
x = +-π/6 + A/2 + πn,n∈Z
x = +-π/6 + 1/2 arcsin 3/5 + πn,∈Z
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Answers & Comments
3/5 sin2x + 4/5 cos 2x = 1
3/5=sinA, 4/5=cosA
sinA*sin 2x + cosA*cos 2x = 1
cos(2x - A) = 1
2x - A = +-arccos 1/2 + 2πn,n∈Z
2x - A = +-π/3 + 2πn,n∈Z
x = +-π/6 + A/2 + πn,n∈Z
x = +-π/6 + 1/2 arcsin 3/5 + πn,∈Z