$\frac{π}{2}$ < θ < π এবং 5sin²θ + 3cos²θ = 4 হলে sin2θ এবং sin3θ এর মান হবে যথাক্রমে - 

$\begin{array}{l}\text {}: 5 \sin ^{2} \theta+3 \cos ^{2} \theta=4 \\ \Rightarrow 5 \sin ^{2} \theta+3 \cos ^{2} \theta=4\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ \Rightarrow \sin ^{2} \theta=\cos ^{2} \theta \Rightarrow \tan ^{2} \theta=1 \\ \Rightarrow \theta=45^{\circ} \cdot \frac{\pi}{2}<\theta<\pi \text { এর জন্য } ; \theta=135^{\circ} \\ \therefore \sin 2 \theta=\sin 270^{\circ}=-1, \sin 3 \theta=\frac{1}{\sqrt{2}}\end{array}$