Question

How do you find all `sin^2 3x-sin^2x=0` in the interval `[0,2pi)`?

Answer 1

The solutions are `S={0,1/4pi,1/2pi,3/4pi,pi,5/4pi,3/2pi,7/4pi}`

Explanation:

We need

`sin^2x+cos^2x=1`

`sin2x=2sinxcox`

`cos2x=1-2sin^2x`

We start by calculating `sin3x`

`sin3x=sin(2x+x)`

`=sin2xcosx+cos2xsinx`

`=2sinxcos^2x+sinx(1-2sin^2x)`

`=2sinx(1-sin^2x)+sinx-2sin^3x`

`=2sinx-2sin^3x+sinx-2sin^3x`

`=3sinx-4sin^3x`

Therefore,

`sin^2(3x)-sin^2x=0`

`(3sinx-4sin^3x)^2-sin^2x=0`

`9sin^2x-24sin^4x+16sin^6x-sin^2x=0`

`16sin^6x-24sin^4x+8sin^2x=0`

`8sin^2x(2sin^4x-3sin^2x+1)=0`

`8sin^2x(2sin^2x-1)(sin^2x-1)=0`

`8sin^2x(sqrt2sinx+1)(sqrt2sinx-1)(sinx+1)(sinx-1)=0`

So,

`sin^2x=0`, `=>`, `x=2pin`, and `x=pi+2pin`

`sqrt2sinx+1=0`, `=>`, `sinx=-1/sqrt2`, `=>`, `x=5/4pi+2pin`, and `x=7/4pi+2pin`

`sqrt2sinx-1=0`, `=>`, `sinx=1/sqrt2`, `=>`, `x=1/4pi+2pin` and `x=3/4pi+2pin`

`sinx+1=0`, `=>`, `sinx=-1`, `=>`, `x=3/2pi+2pin`

`sinx-1=0`, `=>`, `sinx=1`, `=>`, `x=1/2pi+2pin`

Answer 2

`0; pi/4; pi/2; pi`

Explanation:

`f(x) = sin^2 3x - sin^2 x = (sin 3x - sin x)(sin 3x + sin x)`
Apply the 2 trig identities:
`sin a - sin b = 2cos((a + b)/2)sin ((a - b)/2)`
`sin a + sin b = 2sin ((a + b)/2)cos ((a - b)/2)`
In this case -->
f(x) = (2cos 2x.sin x)(2sin 2x.cos x) = 0
f(x) = (2sin 2x.cos 2x)(2sin x.cos x) = sin 4x.sin 2x = 0
Either factor must be zero.

a. sin 2x = 0 --> 2x = 0; `2x = pi`, and `2x = 2pi` -->
`x = 0; x = pi/2`; and `x = pi`
b. sin 4x = 0 --> 4x = 0; `4x = pi`; and `4x = 2pi`-->
`x = 0; x = pi/4`; and `x = pi/2`
Answers in interval `(0, 2pi)`:
`0; pi/4; pi/2; pi`
Check by calculator:
`x = pi/4` --> `sin x = 1/sqrt2` --> `sin^2 x = 1/2` -->
`sin 3x = sin ((3pi)/4) = sqrt2/2` --> `sin^2 ((3pi)/4) = 2/4 = 1/2`.
`sin^2 3x - sin^2 x = 1/2 - 1/2 = 0` Proved.