Mathematics, 19.08.2019 07:00 mine9226

Solve the equation on the interval [0,2pi).
2sin^2x=-3sinx+5

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Solve the equation on the interval [0,2pi).
2sin^2x=-3sinx+5...

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Solve the equation on the interval [0,2pi). 2sin^2x=-3sinx+5

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Another question on Mathematics

Solve the equation on the interval [0,2pi).
2sin^2x=-3sinx+5