solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$ $$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$
I think to do :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$
but i dont get anything.
Or to divied by $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$

$$\sqrt{3} \sin x - \cos x = \sqrt{2}$$
Dividing both sides by 2, we get
$$\frac{\sqrt{3}}{2}\sin x -  \frac{1}{2}\cos x = \frac{\sqrt{2}}{2}$$
By substituting $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$  & $\sin 30^{\circ} = \frac{1}{2}$, we get
$$\sin x \cos30^{\circ}  - \cos x \sin 30^{\circ} = \frac{\sqrt{2}}{2}$$
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\sin 45^{\circ} = \frac{\sqrt{2}}{2}$,
$$\begin{align*}\sin(x - 30^{\circ}) &= \sin 45^{\circ} \\ x - 30^{\circ} &= 45^{\circ} \\ x &= 75^{\circ}\end{align*}.$$
Also, we know that $\sin 135^{\circ} = \frac{\sqrt{2}}{2}$. Then,
$$\begin{align*}\sin(x - 30^{\circ}) &= \sin 135^{\circ} \\ x - 30^{\circ} &= 135^{\circ} \\ x &= 165^{\circ}\end{align*}.$$
Therefore, $x = 30^{\circ}$ and $x = 135^{\circ}$.
