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 $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$
 A: Hint:
$$\sqrt3\sin x-\cos x=\sqrt2\iff \sin\frac\pi3\sin x-\cos\frac\pi3\cos x=\frac{\sqrt2}2$$
A: One of the R-Formulas, a set of formulas for combining such trigonometric expressions, says that
$$a\sin{x} - b\cos{x} = R\sin(x - \alpha)$$
where
$$R = \sqrt{a^2 + b^2}, \alpha = \tan^{-1}{\frac{b}{a}}$$
However, I doubt this solution expresses any room for creativity for this question specifically, as noted by Don Antonio's interesting observation.
A: A difficulty of this equation is that it contains both $\sin(x)$ and $\cos(x)$. You can use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to express $\sin(x) = \pm \sqrt{1 - \cos^2(x)}$ (or $\cos(x)$ as a function of $\sin(x)$), but it is very unwieldy because it introduces square roots (another difficulty) and you need to distinguish the intervals where $\sin(x)$ is positive or negative. It is simpler if you can recognize another trigonometric identity.
The equation looks a bit like $\cos(u) \sin(x) - \sin(u) \cos(x) = a$ for some values $u$ and $a$ to be determined, only with a multiplicative factor. If you had this, you could apply the sine-difference formula: the equation is equivalent to $\sin(x-u) = a$. Your idea to multiply by a constant was on the right; the other part of the puzzle is this identity which guides you towards a multiplicative constant that helps. The equation constrains $\dfrac{\cos(u)}{\sin(u)} = \dfrac{\sqrt 3}{1}$ which you should recognize as having the solution $u = \frac{\pi}{6}$. Since $\cos(\frac{\pi}{6}) = \frac{\sqrt 3}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$, multiply the original equation by $\frac{1}{2}$ to get
$$ \sin\left(x - \frac{\pi}{6}\right) = \frac{\sqrt 2}{2} $$
Alternatively, to find the multiplicative coefficient, you can remark that $\cos^2(u) + \sin^2(u) = 1$, while here you have $(\sqrt 3)^2 + (1)^2 = 4$. Thus you need to divide the equation by $\sqrt 4$ to get coefficients that are a (cos, sin) pair.
(Here I am essentially deriving the formula shown by Yiyuan Lee from a more common identity.)
Since $\frac{\sqrt 2}{2} = \sin(\frac{\pi}{4})$, the equation is equivalent to
$$
x - \frac{\pi}{6} = \frac{\pi}{4} + 2k\pi
\qquad\text{or}\qquad
x - \frac{\pi}{6} = \pi - \frac{\pi}{4} + 2k\pi
$$
i.e. $x = \frac{5\pi}{12} + 2k\pi$ or $x = \frac{11\pi}{12} + 2k\pi$.
A: HINT:
Weierstrass substitution is not a bad alternative either.
It will leave a Quadratic Equation in $\displaystyle\tan\frac x2$ on substitution  and rearrangement of the given relation.
A: $$\sqrt{3} \sin x - \cos x = \sqrt{2}$$
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 $x = 135^{\circ}$.
