that is not an answer. It is a restatement of the problem, using only sines.
2sin^2x-sinx-1 == 0
(2sinx+1)(sinx-1) = 0
sinx = -1/2 or 1, so
x = π/2, 4π/3, 5π/3
http://www.wolframalpha.com/input/?i=cos2x%2Bsinx%3D0+for+x%3D0..2pi
Hello! Can someone please check and see if I did this right? Thanks! :)
Directions: Find the exact solutions of the equation in the interval [0,2pi]
cos2x+sinx=0
My answer:
cos2x+sinx=cos^2x-sin^2x+sinx
=1-sin^2x-sin^2x+sinx
=-2sin^2x+sinx+1=0
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