Question 1090692: Please help me solve/simplify this. Thanks.
Solve:
2sin^2(theta) - cos(theta) - 1 = 0, on [0,2pi]
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! For simplicity, we will replace theta with x.
Using the trig identity, sin^2(x) = 1- cos^2(x), we have
2(1-cos^2(x)) - cos(x) + 1 = 0
2cos^2(x) + cos(x) - 1 = 0
which factors as:
(2cos(x)-1)(cos(x)+1) = 0
Thus cos(x) = -1 and cos(x) = 1/2
If cos(x) = -1 -> x = pi
If cos(x) = -1/2 -> x = 2pi/3 and 4pi/3 on the interval [0,2pi]
|
|
|
