Question

`4 sin x - 3 cos x = sin 3x` ?

Answer 1

Hmmm...

Explanation:

It looks to me like you have slightly misremembered a standard trigonometric identity:

`sin 3x = 3 sin x - 4sin^3 x`

or perhaps:

`cos 3x = 4 cos^3 x - 3 cos x`

We can derive both of these formulae from a combination of de Moivre's theorem and Pythagoras:

`cos nx + i sin nx = (cos x + i sin x)^n`

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

So we find:

`cos 3x+i sin 3x`

`=(cos x + i sin x)^3`

`=cos^3 x + 3 i cos^2 x sin x - 3 cos x sin^2 x - i sin^3 x`

`= (cos^3 x - 3 cos x sin^2 x) + i(3cos^2 x sin x - sin^3 x)`

Equating Real parts we have:

`cos 3x = cos^3 x - 3 cos x sin^2 x`

`color(white)(cos 3x) = cos^3 x - 3 cos x (1 - cos^2 x)`

`color(white)(cos 3x) = 4 cos^3 x - 3 cos x`

Equating imaginary parts we have:

`sin 3x = 3cos^2 x sin x - sin^3 x`

`color(white)(sin 3x) = 3(1 - sin^2 x) sin x - sin^3 x`

`color(white)(sin 3x) = 3sin x - 4sin^3 x`

Answer 2

`x = pi/4 + npi" "` for any integer `n`

Explanation:

Perhaps this is not an attempt at an identity, but an equation to be solved.

In which case:

`4 sin x - 3 cos x = sin 3x = 3 sin x - 4 sin^3 x`

Hence:

`4 sin^3 x + sin x = 3 cos x`

Square both sides to get:

`16 sin^6 x + 8 sin^4 x + sin^2x = 9 cos^2 x = 9 - 9 sin^2 x`

Hence:

`0 = 16 sin^6 x + 8 sin^4 x + 10 sin^2 x - 9`

`color(white)(0) = (2 sin^2 x - 1)(8 sin^4 x + 8 sin^2 x + 9)`

This has only one real root and hence we find:

`sin^2 x = 1/2`

So:

`sin x = +-sqrt(2)/2`

If `sin x = sqrt(2)/2` then:

`3 cos x = 4 sin^3 x + sin x = 3(sqrt(2)/2)`

hence `x = pi/4+2npi" "` for any integer `n`

If `sin x = -sqrt(2)/2` then:

`3 cos x = 4 sin^3 x + sin x = 3(-sqrt(2)/2)`

hence `x = (3pi)/4 + 2npi" "` for any integer `n`

So the general solution is:

`x = pi/4 + npi" "` for any integer `n`