Question

How do you solve `2cos^2x-cosx=3` in the interval `0<=x<=2pi`?

Answer 1

`(2k + 1)pi`

Explanation:

Bring the equation to standard form:
`2cos^2 x - cos x - 3 = 0`
Solve this quadratic equation for cos x.
Since a - b + c = 0, use shortcut.
The 2 real roots are: cos x = - 1 and `cos x = -c/a = 3/2` (rejected as > 1)
Unit circle -->
cos x = -1 --> `x = pi + 2kpi = (2k + 1)pi`