SOLUTION: Solve the equation, where 0° ≤ x < 360°. 2 sin^2 x = 1 − cos x

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the equation, where 0° ≤ x < 360°. 2 sin^2 x = 1 − cos x      Log On


   

Question 1178161: Solve the equation, where 0° ≤ x < 360°.
2 sin^2 x = 1 − cos x
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
sin^2 x=1-cos^2 x
so 2-2cos^2x=1-cos x
0=2 cos^2 x-cos x -1
0=(2 cos x+1)(cos x-1)
cos x =-0.5
cos x=+1
x=0 degrees, 120 degrees, 240 degrees
graph%28300%2C300%2C-6%2C6%2C-1%2C2%2C2sin%28x%29%5E2%2Bcos%28x%29-1%29