SOLUTION: Find all solutions in the interval [0,2pi): 2sin^2(x)=sinx

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Question 737715: Find all solutions in the interval [0,2pi): 2sin^2(x)=sinx
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
2sin^2(x)=sinx
2sin^2(x)-sinx=0
sinx(2sinx-1)=0
either sin x = 0 OR 2sinx-1=0
2sinx-1=0
sinx=1/2
sinx = 1/2 for pi/6 radians and co terminal of pi/6
x=0,pi/6,5pi/6
pi not included