Question

How do you find all `cos2x-cos6x=0` in the interval `[0,2pi)`?

Answer 1

`0, pi/4, pi/2, (3pi)/4, pi, (5pi)/4, (3pi)/2, (7pi)/4`

Explanation:

Use trig identity:
`cos a - cos b = - 2sin ((a+b)/2)sin ((a - b)/2)`
In this case:
`(a + b)/2 = 4x`
`(a - b)/2 = 2x`
`cos (2x) - cos (6x) = 2sin (4x).sin (2x) = 0`
a. `sin 2x = 0` --> Unit circle gives --> `2x = kpi`
`x = (kpi)/2`
b. `sin 4x = 0->` Unit circle --> `4x = kpi`
`x = (kpi)/4`
Answers for `(0, 2pi)`:
`0, pi/4, pi/2, (3pi)/4, pi, (5pi)/4, (3pi)/2, (7pi)/4`