How do you find all values of x in the interval [0, 2pi] in the equation (cos x+1) (2cos^2 x-3 cos x-2) =0?

1 Answer
Nghi N.
Mar 15, 2016

(2pi)/3, pi, (4pi)/3

Explanation:

Solve the 2 factors:
a. (cos x + 1) = 0 --> cos x = -1 --> x = pi
b. 2cos^2 x - 3cos x - 2 = 0.
Use the improved quadratic formula to solve for cos x.
D = d^2 = b^2 - 4ac = 9 + 16 = 25 --> d = +- 5
cos x = -b/(2a) +- d/(2a) = 3/4 +- 5/4.
cos x = 8/4 = 2 (rejected since > 1), and cos x = - 2/4 = -1/2
cos x = - 1/2 --> x = +- (2pi)/3 -->
x = (2pi)/3, and x = (4pi)/3 (co-terminal to -(2pi)/3)
Answers for (0, 2pi):
(2pi)/3, pi, (4pi)/3

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