Question

How do you solve `cos (theta/2) - cos theta = 1` over the interval 0 to 2pi?

Answer 1

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

Explanation:

Use the trig identity to transform cos t to cos (t/2) -->
`cos t = 2cos^2 (t/2) - 1.`
The equation becomes:
`cos (t/2) - 2cos^2 (t/2) + 1 = 1`
`cos (t/2) - 2cos^2 (t/2) = 0`
Put `cos (t/2)` into common factor -->
`cos (t/2)(1 - 2cos (t/2)) = 0`
Trig table and unit circle -->
a. `cos (t/2) = 0` --> There are 2 solution arcs:
`(t/2) = pi/2` --> `t = (2pi)/2 = pi`
`t/2 = (3pi)/2` -->`t = (6pi)/2 = 3pi = pi`
b. `cos (t/2) = 1/2` --> `t/2 = +- pi/3`-->
`t/2 = pi/3` --> `t = (2pi)/3`
and `t/2 = - pi/3` or `t/2 = (5pi)/3` (co-terminal of `-(pi/3))`-->
`t = (10pi)/3 = ((4pi)/3 + 2pi) = (4pi)/3`

Answers for `(0, 2pi):`
`pi, (2pi)/3, (4pi)/3`
Checking these answers by calculator is advised.