# solve a trigonometric equation sqrt(3) sin(x)−cos(x)=sqrt(2)

$\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2}$
I think to do :
$\frac{\left(\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2}\right)}{\sqrt{2}}$
but i dont get anything. Or to divied by $\sqrt{3}$ :
$\frac{\left(\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2}\right)}{\sqrt{3}}$
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barquegese2
Hint:
$\sqrt{3}\mathrm{sin}x-\mathrm{cos}x=\sqrt{2}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}\frac{\pi }{3}\mathrm{sin}x-\mathrm{cos}\frac{\pi }{3}\mathrm{cos}x=\frac{\sqrt{2}}{2}$
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tonan6e
One of the R-Formulas, a set of formulas for combining such trigonometric expressions, says that
$a\mathrm{sin}x-b\mathrm{cos}x=R\mathrm{sin}\left(x-\alpha \right)$
where
$R=\sqrt{{a}^{2}+{b}^{2}},\alpha ={\mathrm{tan}}^{-1}\frac{b}{a}$
However, I doubt this solution expresses any room for creativity for this question specifically, as noted by first answer