Determine sinus and cosinus : $2\sin x + 3\cos x = 3$ $$2\sin x + 3\cos x = 3$$
Seemed easy at first but I have no idea how to determine them. I tried replacing the 3 with $3(\sin^2 x + \cos x^2x)$ But it doesnt work. I also tried switching sides from almost everything and no luck. Please help!

\begin{align}
2\sin x + 3\cos x & = \sqrt{2^2+3^2} \left( \frac 2 {\sqrt{2^2+3^2}} \sin x + \frac 3 {\sqrt{2^2+3^2}} \cos x \right) \\[10pt]
& = \sqrt{13} (\cos\alpha\sin x + \sin\alpha \cos x) \\[10pt]
& = \sqrt{13} \,\sin(\alpha+x) = 3. \\[12pt]
\frac 3 {\sqrt{13}} & = \sin(\alpha+x).
\end{align}
Since $3/\sqrt{13} < 1,$ this can be solved.
Since $\sin\alpha/\cos\alpha = 3/2,$ the number $\alpha$ must be a number whose tangent is $3/2$ and whose sine and cosine are both positive, i.e. in the first quadrant.
