Solve $\tan x = \frac{p}{q}$, where $p, q\in\mathbb{Z}$ such that $3\cos x-4\sin x=-5$ A Calculus A level trigonometry problem:

Solve $\tan x = \dfrac{p}{q}$ where $p,q\in\mathbb{Z}$ such that  $$3\cos x\ - 4\sin x = -5$$

I tried moving terms to one side, but that doesn't help much.
Any ideas?
 A: You also need to use $\cos^2x+\sin^2x=1$. The simultaneous equations have solution $\cos x=-\frac35,\,\sin x=\frac45$ (it helps to note $(3,\,-4,\,-5)$ is a Pythagorean triple with some sign changes), so $\tan x=\frac{\sin x}{\cos x}=-\frac43$.
A: Note that $$\cos x=\dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)}\qquad\qquad \sin x=\dfrac{2\tan(x/2)}{1+\tan^2(x/2)}$$ and substitute this in to your eqyation. Then you will get, $$3(1-t^2)-8t+5(1+t^2)=2(t^2-4t+4)=2(t-2)^2=0,$$ where $t=\tan(x/2).$ Then use $$\tan x=\dfrac{2\tan(x/2)}{1-\tan^2(x/2)}.$$ Hope you can take it from here.
A: $$\tan x=\dfrac pq\implies q\sin x-p\cos x=0\ \ \ \ (1)$$
$$3\cos x-4\sin x=-5\ \ \ \ (2)$$
Solve the two simultaneous equations for $\sin x,\cos x$
Use $\cos^2x+\sin^2x=1$ to eliminate $x$
See also: Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$
A: If  $$a\cos x-b\sin x=-\sqrt{a^2+b^2}$$
Let $a=r\cos A,b=r\sin A$ where $r>0$ for $r=0\implies a=b=0$
$$\dfrac{r\sin A}{r\cos A}=?$$
Squaring and adding we have $$r^2=?$$
So, we have $$-r=r(\cos A\cos x-\sin A\sin x)$$
As $r\ne0$ $$\cos(x+A)=-1$$
$$\implies x+A=(2n+1)\pi$$
$$\implies\tan x=\tan((2n+1)\pi-A)=-\tan A=?$$
A: Rewrite $3\cos x\ - 4\sin x = -5$ as $\sin(x-\tan^{-1}\frac34)=1$,
which yields $x=\pi n +\frac\pi2+\tan^{-1}\frac34 $. 
Thus,
$$\tan x = \tan ( \frac\pi2+\tan^{-1}\frac34 )=\cot( -\tan^{-1}\frac34  )
=-\cot( \cot^{-1}\frac43 ) =-\frac43$$
