Question

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

Answer 1

`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)