Question

How do you solve `2cos^2x+cosx -1=0` for `0<=x<=2pi`?

Answer 1

`pi/3; pi; (5pi)/3`

Explanation:

`2cos^2 x + cos x - 1 = 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 = 1/2`
a. cos x = -1
Unit circle gives:
`x = pi`
b. `cos x = 1/2`
Trig table and unit circle give -->
`x = +- pi/3`
Note.
`x = - pi/3` is co-terminal to `x = (5pi)/3`
Answers for `(0, 2pi)`
`pi/3; pi; (5pi)/3`