# Precalculus

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2sin^2x+3cosx=3 find all solutions in the interval [0,2pi]

• Precalculus -

Use the identity sin²(x)=1-cos²(x).

So
2sin^2x+3cosx=3 becomes
2cos²(x)-3cos(x)-1=0

I get cos(x)=1 or cos(x)=1/2

On the interval [0,2π], each value of cos(x) has two solutions for a total of 4 values of x.

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