Question

Solve over the interval [0, 360):
2sin^2(x)=3-4cos(x)

Currently stuck up to: 2cos^2(x)-4cos(x)+1=0

Answer 1

hint: sin^2x = 1 - cos^2x

Answer 2

Right, I was able to approach the problem using the Pyth Ident until I got stuck at:

Answer 3

2cos^2(x)-4cos(x)+1=0

Answer 4

Can you solve this? \(2z^{2} - 4z + 1 = 0\)