Question

How do you solve the equation on the interval [0,2pi) for `2(cos(t))^2-cos(t)-1=0`?

Answer 1

`0, (2pi)/3, (4pi)/3, 2pi`

Explanation:

Solve: `2cos^2 t - cos t - 1 = 0` for cos t.
Since a + b + c = 0, use shortcut. The 2 real roots are: cos t = 1 and `cos t = c/a = - 1/2`.
a. cos t = 1 --> t = 0 or t = `2pi`
b. `cos t = - 1/2`
Trig table and unit circle -->
`cos t = - 1/2` --> `t = +- (2pi)/3`
Arc `(4pi)/3` is co-terminal to arc `(-2pi/3)`.
Answer for `(0, 2pi):`
`0, (2pi)/3, (4pi)/3, 2pi`