SOLUTION: 2sin^2x-3sinx+1=0, 0 equal to or less than x <2pi

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Question 1007573: 2sin^2x-3sinx+1=0, 0 equal to or less than x <2pi
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
From 2sin^2x-3sinx+1=0, we factor and get
(2 sin x - 1)(sin x - 1) = 0 so that
sin x = 1/2 and sin x = 1 so that
x = pi/6 and 5pi/ 6 and pi/2