Question

How do you solve the equation on the interval [0,2pi) for `cos^2 theta - cos theta - 2 = 0`?

Answer 1

`t = pi`

Explanation:

f(t) = cos^2 t - cos t - 2 = 0
This is a quadratic equation for cos t.
Since a - b + c = 0, use shortcut. Two real roots: cos t = - 1 and
cos t = -c/a = 2
a. cos t = - 1 --> `t = pi`
b. cos t = 2 (rejected since > 1)
Answer for `(0, 2pi)`
`t = pi`
Check
`t = pi` --> `cos ^2 t = 1` --> cos t = -1.
`cos^2 t - cos t - 2 = 1 - (-1) - 2 = 0`. OK