How do you solve 2cos2 x - 3 cos x + 1 = 02cos2x−3cosx+1=0 over the interval 0 to 2pi?

1 Answer
Nghi N.
Feb 5, 2016

0, 360, 104^@48, 255^@520,360,104∘48,255∘52

Explanation:

Use trig identity: cos 2x = 2cos^2 x - 1.cos2x=2cos2x−1.
Replace in the equation cos 2x by (2cos^2 x - 1)(2cos2x−1), we get:
2(2cos^2 x - 1) - 3cos x + 1 = 02(2cos2x−1)−3cosx+1=0
4cos^2 x - 3cos x - 1 = 04cos2x−3cosx−1=0. Solve this quadratic equation.
Since a + b + c = 0, use shortcut. one real root is cos x = 1 and the other is cos x = c/a = -1/4.cosx=ca=−14.
cos x = 1 --> x = 0 and x = pi.
cos x = -1/4cosx=−14 --> x = +- 104.48x=±104.48
Answers: 0, 360^@, 104^@48 , 255^@520,360∘,104∘48,255∘52 (or -104.48)

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