$2 \left ( \sin{θ} \cos{θ} + \sqrt{3} \right ) = \sqrt{3} \cos{θ} + 4 \sin{θ}$  এবং 0<θ<300° হলে θ=কত? 

$2(\sin \theta \cdot \cos \theta+\sqrt{3})=\sqrt{3} \cos \theta+4 \sin \theta$

বা, $2 \sin \theta \cdot \cos \theta-4 \sin \theta=\sqrt{3} \cos \theta-2 \sqrt{3}$

বা$2 \sin \theta \cdot(\cos \theta-2)=\sqrt{3}(\cos \theta-2)$

বা, $2 \sin \theta=\sqrt{3} \therefore \sin \theta=\frac{\sqrt{3}}{2} \therefore \theta=60^{\circ}, 120^{\circ}$