Question

How do you solve `2cos^2 theta + cos theta -1 = 0`?

Answer 1

`(1 + 2k)pi; pi/3 + 2kpi; (5pi)/3 + 2kpi`

Explanation:

Solve the quadratic equation for cos t.
`f(t) = 2cos^2 t + cos t - 1 = 0`
Since a - b + c = 0, use shortcut. The 2 real roots are:
cos t = -1 and `cos t = -c/a = 1/2`
Use trig table of special arcs and unit circle -->

a. cos t = -1 --> `t = pi + 2kpi`
General answers: `t = (1 + 2k)pi`
b. `cos t = 1/2` -->`t = +- pi/3 + 2kpi`
arc `(-pi)/3` and arc `(5pi)/3` are co-terminal.
General answers:
`(1 + 2k)pi`
`pi/3 + 2kpi`
`(5pi)/3 + 2kpi`