সমাধান কর:

$2 \left( \sin{x} \cos{x} + \sqrt{3} \right ) = \sqrt{3} \cos{x} + 4 \sin{x} , 0$

$2 \sin x \cos x+2\sqrt{3}-\sqrt{3}\cos x-4\sin x=0$

$\Rightarrow 2 \sin x (\cos x-2)-\sqrt{3}(\cos x-2)=0$$\Rightarrow (2 \sin x-\sqrt{3})(\cos x-2)=0$

$\therefore 2\sin x-\sqrt{3}=0 or \cos x-2=0; but \cos x-2\ne 0$

$\therefore \sin x=\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3} \therefore x=n\pi +{(-1)}^n\frac{\pi }{3} [n\in Z]$

$n=0$হলে,$x=\frac{\pi }{3}; n=1$হলে,$x=\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

$x=2$হলে, $x=2\pi +\frac{\pi }{3}=\frac{7\pi }{3}$$\therefore$নির্ণেয় সমাধান$=\left\{\frac{\pi }{3}, \frac{2\pi }{3}\right\}$