$3 sin^2 \theta + 2 sin^2 \phi = 1$
$\Rightarrow 3 sin^2 \theta = 1 - 2 sin^2 \phi \ \ \ \ \ ............(1)$
$\Rightarrow 3 sin^2 \theta = cos2 \phi \ \ \ \ \ \ \Rightarrow 3sin2\theta = 2sin2\phi \ \ \ \ \ .........(2)$
અહી, $3 sin\theta cos \theta = sin2\phi$
વર્ગ અને સરવાળો લેતાં (સમી $1$ + સમી $2$ પરથી)
$\Rightarrow 9 sin^2 \theta (sin^2 \theta + cos^2 \theta) = 1$
$\Rightarrow 9 sin^2 \theta = 1 \ \ \ \ \ \ \ \Rightarrow sin^2\theta = \frac{1}{9}$
$\Rightarrow sin \theta = \frac{1}{3}, cos = \frac{2\sqrt{2}}{3}$
$\therefore cos 2 \phi = 3 \left(\frac{1}{9}\right) = \frac{1}{3}$
$\therefore sin 2 \phi = \frac{2\sqrt{2}}{3}$
$\therefore cos (\theta + 2\phi) = cos \theta cos 2\phi - sin \theta sin 2\phi$
$= \frac{2 \sqrt{2}}{3} \left(\frac{1}{3}\right) - \frac{1}{3} \left(\frac{2\sqrt{2}}{3}\right) = 0 $
$\Rightarrow \theta + 2\phi = \frac{\pi}{2}$
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