Find all solutions to 2sin^2x-3cosx=3.

Answer 1

If you rewrite this as 2(1 - cos²x-3cosx-3=0 you get 2cos²x + 3cosx + 1 = 0

If you let cosx = z,

Then, 2z² + 3z + 1 = 0 using the quadratic equation for z, we find that

z = -3/4 +- 1/4 or -1/2 and -1 So, wherever cosx = -1/2 or -1 are all solutions

This starts with x = 120, 180, & 240 degrees and all of the 2pi additions to + & - infinity.