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 
$$\dfrac{d}{dx} 5\cos(x)-12\sin (x) = 13  = 0$$
 A: HINT
For these kind of equations we can use by $t=\tan \frac {x} 2$ with $\frac {x} 2 \neq\frac{\pi}2+k\pi$ (which in that case are not solutions) the following identities


*

*$\sin (x) =\frac{2t}{1+t^2}$

*$\cos (x) =\frac{1-t^2}{1+t^2}$
to obtain a quadratic equation in $t$ that is
$$5\frac{1-t^2}{1+t^2}-12\frac{2t}{1+t^2}=13 \iff 9t^2+12t+4=(3t+2)^2=0$$
As an alternative we can also use the following


*

*$a\sin x+b\cos x=c\sin(x+\varphi)$

*$c = \sqrt{a^2 + b^2}$

*$\varphi = \operatorname{atan2} \left( b, a \right)$

A: 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!
A: Hint
By Cauchy Schwarz you have
$$169=13^2=(5\cos(x)-12\sin (x))^2 \leq (5^2+12^2)(\cos^2(x)+\sin^2(x))=169$$
Therefore, you must have equality in CS, and hence 
$$\frac{\cos(x)}{5}=\frac{\sin(x)}{-12}$$
Now combine this equality with $$\sin^2(x)+\cos^2(x)=1$$
A: Hint: the left-hand side is $13\cos (x+\arctan\frac{12}{5})$, so you have to set a cosine to $1$.
A: The expression $a\cos x+ b\sin x$ can be rewritten as one of 
\begin{cases} a\cos x+ b\sin x=r\cos(x-\varphi)\\a \sin x + b \cos x = r\sin(x+\varphi) \end{cases}
where $r=\sqrt{a^2+b^2}$, $\varphi=\arctan(b/a)$ if $a>0$, $\varphi=\pi+\arctan(b/a)$ if $a<0$
We use the first equation, $a=5, b=-12$:
$$\sqrt{5^2+12^2}\cos(x-\arctan(-\frac{12}{5}))=13$$
Divide both sides by 13
$$\cos(x-\arctan(-\frac{12}{5}))=1$$
The cos function is only 1 for $n2\pi$, $\arctan(-u)=-\arctan(u)$
$$x+\arctan(12/5)=n2\pi,\quad n\in \mathbb{z}$$
$$\boxed{x=n2\pi-\arctan(\frac{12}{5}),\quad n\in \mathbb{z}}$$
