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!
 A: HINT
You have two equations $$ 2\sin(x) + 3\cos(x) = 3\\\sin(x)^2+\cos(x)^2 = 1.$$
Maybe for clarity, replace $\sin$ with $y$ and $\cos$ with $x$ so you have $$ 2y+3x = 3 \\x ^2 + y^2=1.$$
One of these things is a line and one is a circle... find where they intersect. (Further hint: one of these is trivial and one less so.) 
This will give the possible values for $\sin(x)$ and $\cos(x)$ which is what you asked for, but if you want the possible values of $x$ there will be an infinite number corresponding to each of the above solutions due to periodicity.
A: \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.
A: Hint:
You can rewrite the equation as
\begin{align}
2\sin x +3\cos x=3&\iff 2\sin x=3(1-\cos x)\iff 4\sin\dfrac x2\cos\dfrac x2=6\sin^2\dfrac x2\\[1ex]
&\iff\begin{cases}
\sin \dfrac x2=0\\\;\text{ or}\\[-2ex]2\cos\dfrac x2=3\sin \dfrac x2
\end{cases}\iff\begin{cases}
\sin \dfrac x2=0\\\;\text{ or}\\[-2ex]\tan\dfrac x2=\dfrac23
\end{cases}\\&\iff\begin{cases}
\dfrac x2\equiv 0 \pmod\pi\\\;\text{ or}\\[-2ex]\dfrac x2\equiv\arctan\dfrac23\pmod{\pi}
\end{cases}\iff\begin{cases}x\equiv 0 \pmod{2\pi}\\\;\text{ or}\\x\equiv 2\arctan\dfrac23\pmod{2\pi}\end{cases}
\end{align}
