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

1 Answer

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