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

1 Answer
Nghi N.
Feb 5, 2016

#0, 360, 104^@48, 255^@52#

Explanation:

Use trig identity: #cos 2x = 2cos^2 x - 1.#
Replace in the equation cos 2x by #(2cos^2 x - 1)#, we get:
#2(2cos^2 x - 1) - 3cos x + 1 = 0#
#4cos^2 x - 3cos x - 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.#
cos x = 1 --> x = 0 and x = pi.
#cos x = -1/4# --> #x = +- 104.48#
Answers: #0, 360^@, 104^@48 , 255^@52# (or -104.48)

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