Question

The number of solutions of $\sin \left(\frac{\pi x}{2\sqrt{3}}\right)=x^2-2\sqrt{3}x+4$ is

Answer 1

$\sin \left(\frac{\pi x}{2\sqrt{3}}\right)=x^2-2\sqrt{3}x+4\Rightarrow \sin \left(\frac{\pi x}{2\sqrt{3}}\right)=x^2-2\sqrt{3}x+(\sqrt{3}{)}^2+1\Rightarrow \sin \left(\frac{\pi x}{2\sqrt{3}}\right)=(x-\sqrt{3}{)}^2+1$
$\text{L.H.S.}\le 1$ and $\text{R.H.S.}\ge 1$
So, solution exist only when
$\text{L.H.S.}=\text{R.H.S.}=1$
$\Rightarrow \text{R.H.S.}=1$
$\Rightarrow (x-\sqrt{3}{)}^2+1=1$
$\Rightarrow x=\sqrt{3}$

At $x=\sqrt{3},$
$\text{L.H.S.}=\sin \left(\frac{\pi \sqrt{3}}{2\sqrt{3}}\right)=1$

Therefore, $x=\sqrt{3}$ is the only solution.