Question 677925: Solve for x in [0,2pi]:
cos 2x = -sin x
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve for x in [0,2pi]:
cos 2x = -sin x
cos^2x-sin^2x=-sinx
1-sin^2x-sin^2x=-sinx
-2sin^2x+sinx+1=0
2sin^2x-sinx-1=0
(2sinx+1)(sinx-1)=0
..
2sinx+1=0
sinx=-1/2
x=7π/6, 11π/6 (in quadrants III and IV where sin<0)
and
sinx-1=0
sinx=1
x=π/2
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