If $3\sin ^{2}\theta +2\sin ^{2}\phi=0$ and $3\sin 2\theta +2\sin 2\phi=0$, $0< \theta < \frac{\pi }{2}$ and $0< \phi < \frac{\pi }{2}$, then the value of $\theta +2\phi$

Griem:

$\begin{aligned} & 3 \sin ^{2} \theta+2 \sin ^{2} \phi=1 \\ \Rightarrow & 3 \sin ^{2} \theta=1-2 \sin ^{2} \phi \\ \Rightarrow & 3 \sin ^{2} \theta=\cos 2 \phi…….(1) \end{aligned}$

and $3 \sin \theta \cos \theta=\sin 2 \phi$……….(2)

On squaring and adding Equations (1) and (2), nee get

$\begin{array}{l} 9 \sin ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=1 \\ \sin \theta=\frac{1}{3} \text { and } \cos \theta=\frac{2 \sqrt{2}}{3} \end{array}$

$\sin \theta=\frac{1}{3}$ and $\cos \theta=\frac{2 \sqrt{2}}{3}$

$\therefore \cos 2 \phi=3 \times \frac{1}{9}=1 / 3$ and $\sin 2 \phi=\frac{2 \sqrt{2}}{3}$

Now, $\cos (\theta+2 \phi)=\cos \theta \cos 2 \phi-\sin \theta \sin 2 \phi$

$=\frac{2 \sqrt{2}}{3} \cdot \frac{1}{3}-\frac{1}{3} \cdot \frac{2 \sqrt{2}}{3}=0$

and $\theta+2 \phi<\frac{3 \pi}{2}$

$\therefore \theta+2 \phi=\pi / 2$