Posted by **Megan** on Wednesday, December 8, 2010 at 8:20pm.

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

- Precalculus -
**MathMate**, Wednesday, December 8, 2010 at 9:32pm
Use the identity sin²(x)=1-cos²(x).

So

2sin^2x+3cosx=3 becomes

2cos²(x)-3cos(x)-1=0

Solve the quadratic for cos(x),

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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