Precalculus
posted by Megan .
2sin^2x+3cosx=3 find all solutions in the interval [0,2pi]

Use the identity sin²(x)=1cos²(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.