Solve: $\displaystyle 5\cos(x)+12\sin(x)=3$

Let $\displaystyle t=\tan\left(\frac{x}{2}\right)$ therefore we can write the equation as:

$\displaystyle 5\cdot \left(\frac{1-t^2}{1+t^2}\right)+12\cdot \sin\left(\frac{2t}{1+t^2}\right)-3=0$

$\displaystyle \Leftrightarrow \frac{5(1-t^2)+24t-3(1+t^2)}{1+t^2}=0$

$\displaystyle \Leftrightarrow 5(1-t^2)+24t-3(1+t^2)=0$

$\displaystyle \Leftrightarrow 5-5t^2+24t-3-3t^2=0$

$\displaystyle \Leftrightarrow -8t^2+24t+2=0$

Solutions:

$\displaystyle t_1=\frac{3-\sqrt{10}}{2}$ and $\displaystyle t_2=\frac{3+\sqrt{10}}{2}$

If we do the back-substitution:

$\displaystyle x=2\arctan\left(\frac{3+\sqrt{10}}{2}\right)+2k\pi$

$\displaystyle x=2\arctan\left(\frac{3-\sqrt{10}}{2}\right)+2k\pi$

(and $\displaystyle k \in \mathbb{Z}$)

So if we define the solution set:

$\displaystyle V=\{2\arctan\left(\frac{3-\sqrt{10}}{2}\right)+2k\pi,2\arctan\left(\frac{3+\ sqrt{10}}{2}\right)+2k\pi\}$

So we've solved the equation now there's nothing more to do and you don't have to worry about in which quadrant the function appears because that's the reason we define an infinite set of solutions.

If you want to be sure about the solution then take a look at wolphram alpha to compare.

Do you become the same result with the method The Chaz has given you?