$\xrightarrow[{{R_{31}}( - 1)}]{{{R_{32}}( - 1)}}$
$\begin{vmatrix}1&0&-1\\0&1&-1\\\sin^2\theta&\cos^2\theta&1+4\sin4\theta\end{vmatrix}=0$
$\xrightarrow[{{C_{13}}( 1)}]{{{R_{23}}( 1)}}$
$\begin{vmatrix}1&0&-1\\0&1&-1\\ \sin^2\theta&\cos^2\theta&1+\sin^2\theta+\cos^2\theta+4\sin4\theta\end {vmatrix}=0$
$\begin{vmatrix}1&0&-1\\0&1&-1\\\sin^2\theta&\cos^2\theta&2+4\sin4\theta\end{vmatrix}=0$
$\therefore 1[2+4sin \theta -0]=0$
$\therefore sin4 \theta = \frac{-1}{2}$
$\therefore sin4 \theta = sin \frac{- \pi}{6}$
$\therefore 4 \theta =k \pi +(-1)^{k} { \frac{- \pi}{6}} , k \in Z$
$ k \leq { \frac{-\pi}{2}} $ હોવાથી
$4 \theta =k \pi -(-1)^{k} { \frac{ \pi}{6}} , k \in Z$
$ \theta = \frac{k \pi}{4} -(-1)^{k} { \frac{- \pi}{24}} , k \in Z$
$ k=1$ અને $ k=2$ શક્ય બને,
$ k=1$ માટે
$\theta = \frac{\pi}{4}+\frac{\pi}{24}=\frac{7\pi}{24}\in (0,\frac{\pi}{2})$
$ k=2$ માટે
$\theta = \frac{\pi}{2}-\frac{\pi}{24}=\frac{11\pi}{24}\in (0,\frac{\pi}{2})$
ઉકેલગણ $\left\{ {\frac{{7\pi }}{{24}},\frac{{11\pi }}{{24}}} \right\}$
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