You can put this solution on YOUR website! Find all solutions of the equation 2sin^2(x)−cosx=1 in the interval [0,2pi)
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sin^2 = 1-cos^2
--> 2(1-cos^2) - cos - 1 = 0
2 - 2cos^2 - cos - 1 = 0
2cos^2 + cos - 1 = 0
Sub y for cos
2y^2 + y - 1 = 0
(2y - 1)*(y + 1) = 0
y = cos(x) = -1
x = pi
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y = cos(x) = 1/2
x = pi/3, 5pi/3