Find all the solutions in the interval [0,2π)
sin²x = 2-2cosX
1 answer:
Step-by-step explanation:
sin²x = 2 - 2cosx
1 - cos²x = 2 - 2cosx (because sin²x + cos²x = 1)
cos²x - 2cosx + 1 = 0
(cosx - 1)² = 0
cosx - 1 = 0
cosx = 1
Hence x = 0.
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