Question

Find all exact solutions on the interval [0,2pi).
cos(8x)+cos(4x)=0

Answer 1

let's use the identity
cos (2A) = 2cos^2 (A) - 1

cos(8x)+cos(4x)=0
2cos^2 (4x) - 1 + cos(4x) = 0
let cos(4x) = y
then we have
2y^2 + y - 1 = 0
(2y - 1)(y + 1) = 0
y = 1/2 or y = -1

if y = 1/2, cos 4x = 1/2
4x = π/3 or 4x = 2π - π/3 = 5π/3
x = π/12 , or x = 5π/12 , (15° or 75°)

but the period of cos(4x) = 2π/4 = π/2 , (90°)

so adding multiples of π/2 will give us more answers.
<b>x = π/12, 7π/12, 13π/12, 19π/12, </b>
in degrees: 15°, 105°, 195°, 285°, 375° <--- the last one is too big

if y = -1
cos 4x = -1
4x = π
x = π/4 , with the same period of cos(4x) to be π/2
<b>x = π/4, 3π/4, 5π/4, 7π/4</b>
in degrees: 45°, 135°, 225°, 315°