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Find the exact solution of cos2x+sinx=0 in the interval [0,2pi)

1 answer

cos2x+sinx = 0
1 - 2sin^2x + sinx = 0
(2sinx+1)(sinx-1) = 0

sinx = -1/2 in QIII,IV
sinx = 1

I expect you can take it from there.
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