How do you solve `-sqrt(3)cos x + sin x = -sqrt(3)`?

Given that

`-\sqrt3\cos x+\sin x=-\sqrt3`

`\sqrt3\cos x-\sin x=\sqrt3`

`(\sqrt3/2)\cos x-(1/2)\sin x=\sqrt3/2`

`\cos x\cos(\pi/6)-\sin x\sin(\pi/6)=\cos(\pi/6)`

`\cos (x+\pi/6)=\cos(\pi/6)`

`(x+\pi/6)=2n\pi\pm\pi/6`

`x=2n\pi\pm\pi/6-\pi/6`

Where, `n` is any integer i.e. `n=0, \pm1, \pm2, \pm3, \ldots`

Given: `-sqrt(3) cos x + sin x = - sqrt(3)`

Rearrange: `" "sqrt(3) - sqrt(3) cos x + sin x = 0`

When we think of `sqrt(3)` in terms of cosine, realize `cos(pi/6) = sqrt(3)/2`

Group the second two terms and multiply and divide both terms by `2/2`:

`sqrt(3) - 2((sqrt(3))/2 cos x - 1/2sin x) = 0`

Substitute in `cos (pi/6) " for " sqrt(3)/2 " and " sin (pi/6) " for " 1/2:`

`sqrt(3) - 2(cos (pi/6) cos x - sin (pi/6) * sin x) = 0`

Realize that `" "cos(x + pi/6) = cos (pi/6) cos x - sin (pi/6) * sin x`

`sqrt(3) - 2cos(x + pi/6) = 0`

`- 2cos(x + pi/6) = -sqrt(3)`

`cos(x + pi/6) = sqrt(3)/2`

Take the inverse cosine of both sides:

`cos^-1(cos(x + pi/6)) = cos^-1(sqrt(3)/2)`

`x + pi/6 = 2 pi n + pi/6`

and `x + pi/6 = 2 pi n + (11 pi)/6`

Simplify:

`x = 2 pi n " and " x = 2 pi n +(10 pi/6) = 2 pi n + (5 pi)/3`