Question

Solve equation: (Is my answer correct ?)

Answer 1

`x = kpi; x = pi/2 + kpi`
`x = pi/6 + (kpi)/3`

Explanation:

`sin^2 2x = sin^2 3x - sin^2 x`(1)
Develop the right side:
RS = (sin 3x - sin x)(sin 3x + sin x)
RS = (2cos 2x.sin x)(2sin 2x.cos x) = (2sin x.cos x)(2sin 2x.cos 2x)=
RS = sin 2x.sin 4x
Equation (1) becomes:
#sin^2 2x = sin 2x.sin 4x sin^2 2x - sin 2x sin 4x = 0#
`sin 2x(sin 2x - sin 4x) = 0`
Either factor should be zero.

a. sin 2x = 0. Unit circle gives:
`2x = 2kpi`--> `x = kpi`
`2x = pi + 2kpi` --> `x= pi/2 + kpi`

b. `sin 4x = sin 2x`
Unit circle gives:
`4x = 2x + 2kpi`, and `4x = pi - 2x + 2kpi`
1. `4x = 2x + 2kpi` --> `2x = 2kpi` --> `x = kpi`
2. `4x = pi - 2x + 2kpi` --> `6x = pi + 2kpi`
`x = (pi)/6 + (kpi)/3`