Question 1086221: Help please:
2sin^2(theta) - cos(theta) - 1 = 0, on [0,2pi]
Thank you.
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! For simplicity, we let x = theta.
To get an equation in cos(x), use the identity sin^2(x) = 1-cos^2(x)
Then we have
2(1 - cos^2(x)) - cos(x) - 1 = 0
2cos^2(x) + cos(x) - 1 = 0
Factor:
(2cos(x) - 1)(cos(x) + 1) = 0
This has two solutions, cos(x) = -1 and cos(x) = 1/2
I leave it as an exercise to determine the values of x which satisfy the two conditions.
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