SOLUTION: Find all solutions of the equation in the interval [0,2pi). sin^2x=2-2cosx Write your answer in radians in terms of pi. If there is more than one solution, separate them wit

Question 1036282: Find all solutions of the equation in the interval [0,2pi).
sin^2x=2-2cosx
Write your answer in radians in terms of pi.
If there is more than one solution, separate them with commas.
Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
Hi there,
sin^2x = 2 - 2cosx
Using cos^2x + sin^2x = 1
so, sin^2x = 1 - cos^2x
1 - cos^2x = 2 - 2cosx
Rearrange.
cos^2x - 2cosx + 1 = 0
(cosx - 1)(cosx - 1) = 0
cosx - 1 = 0
cosx = 1
x = 0, 2π
Hope this helps :-)
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