Question

Solve equation: (Is my answer correct ?)

`sin^2x + sin^2 2x = sin^2 3x`
`sin^2x + (2sinxcosx)^2 = [sinx(3-4sin^2x)]`
`sin^2x +4sin^2xcos^2x = sin^2x(9-24sin^2x +16sin^4x)`
`sin^2x +4sin^2xcos^2x = 16sin^6x - 24sin^4x +9sin^2x`
`4sin^2x(4sin^4x - 6sin^2x +2 - cos^2x) =0`
`sin^2x = 0`
`x=kπ`
⋁ `4sin^4x - 6sin^2x+2-1+sin^2x=0`
`4sin^4x - 5sin^2x +1 =0`
`sin^2x = t`
`4t^2-5t+1=0`
∆=9
`t_1 = 1/4` and `t_2 =1`

so...

for `t_1 = 1/4` :
`1/2 = sinx` or `-1/2 = sinx`
`x=π/6 + 2kπ or x=-π/6 + 2kπ`

for `t_2 = 1` :
`sinx =1 or sinx =-1`
`x=π/2+2kπ or x=-π/2+2kπ`

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`x=π/6 + kπ ⋁ x=π/2 + kπ`

Answer 1

`x = +- 37^@76 + k180^@`
`x = kpi`, and `x = pi/2 + kpi`

Explanation:

`sin^2 x - sin^2 3x + sin^2 2x = 0`
`(1 - cos 2x) - (1 - cos 6x) + sin^2 2x = 0`
`(cos 6x - cos 2x) + sin^2 2x = 0`
`- 2sin 4x.sin 2x + sin^2 2x = 0`
`sin 2x (sin 2x - 2sin 4x) = sin 2x(sin 2x - 4sin2x.cos 2x) = 0`
`sin^2 2x(1 - 4cos 2x) = 0`
Either factor should be zero
.
a. `sin 2x = 0`. Unit circle gives:
1. `2x = 2kpi` --> `x = kpi`
2. `2x = pi + 2kpi` --> `x = pi/2 + kpi`
b. `4cos 2x = 1` --> `cos 2x = 1/4`
Calculator and unit circle give -->
1. `2x = +- 75^@52 + k360^@` -->
`x = +-37^@76 + k180^@`