Determining the general solution for the trigonometric equation $ 5\cos(x)-12\sin (x) = 13 $ Given that
$$5\cos(x)-12\sin (x) = 13 $$
I'm trying to evaluate the general solution for that expression. It reminds me of $5-12-13$ triangle. Since we don't know the degree of $x$, I couldn't proceed further. Specifically, let's take its derivate, which yields 
$$\dfrac{d}{dx} 5\cos(x)-12\sin (x) = 13  = 0$$

This has a general method: divide the whole equation by $\;\sqrt{5^2+12^2}=13\;$ , so the equation becomes
$$\frac5{13}\cos x-\frac{12}{13}\sin x=1$$
Since $\;\left(\frac5{13}\right)^2+\left(\frac{12}{13}\right)^2=1\;$ , there exists $\;\alpha\in\Bbb R\;$ (in fact, we can choose this value in an infinite number of ways...) such that $\;\cos\alpha=\frac{12}{13}\;,\;\;\sin\alpha=\frac5{13}\;$ , so the equations becomes
$$\sin\alpha\cos x-\sin x\cos\alpha=1\stackrel{\text{trig. identity}}\iff\sin(\alpha-x)=1\ldots$$
Try now to take it from here. And BTW: some high schools specifically forbid to use calculus when solving trigonometric equations!
