Where Did $x$ Go?

Square both equations to give:

$a^2 cos^2{x}+b^2 sin^2{x}+2ab \sin{x} \cos{x}=9 \\ a^2 sin^2{x}+b^2 cos^2{x}-2ab \sin{x} \cos{x}=16$

Adding these two together gives:

$(a^2+b^2)(\sin^2{x}+\cos^2{x})=25$

As $\sin^2{x}+\cos^2{x}=1$ we have:

$\color{#20A900}{\boxed{\boxed{a^2+b^2=25}}}$

Relevant wiki: Fundamental Trigonometric Identities - Problem Solving (Easy)

Squaring $1st$ equation and using $\sin^2x=1-\cos^2x$ :

$a^2(1-\sin^2x)+b^2(1-\cos^2x)+2ab\sin x\cos x=16$ $\implies a^2+b^2-16=(a\sin x-b\cos x)^2=4^2$ $\large\implies a^2+b^2=\color{#0C6AC7}{\boxed{25}}$

Assuming this works for any $x$ , we can simplify things up a bit. For instance, let's change the $x$ to $\frac{\pi}{2}$ : $\begin{cases} a\cos(\frac{\pi}{2}) + b\sin(\frac{\pi}{2}) = 3 \\ a\sin(\frac{\pi}{2}) - b\cos(\frac{\pi}{2}) = 4 \end{cases}$ $= \begin{cases} a\times0 + b\times1 = 3 \\ a\times1 - b\times0 = 4 \end{cases}$ From which we can tell that $a = 4$ and $b = 3$ $\rightarrow a^2 + b^2 = \boxed{25}$