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Let $\theta \in\left(0, \frac{\pi}{2}\right)$. If the system of linear equations
$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$
has a non-trivial solution, then the value of $\theta$ is :
$\frac{4 \pi}{9}$
$\frac{7 \pi}{18}$
$\frac{\pi}{18}$
$\frac{5 \pi}{18}$
Solution
$Case-I$
$\left| {{\mkern 1mu} \begin{array}{*{20}{c}} {1 + {{\cos }^2}\theta }&{{{\sin }^2}\theta }&{4\sin 3\theta }\\ {{{\cos }^2}\theta }&{1 + {{\sin }^2}\theta }&{4\sin 3\theta }\\ {{{\cos }^2}\theta }&{{{\sin }^2}\theta }&{1 + 4\sin 3\theta } \end{array}{\mkern 1mu} } \right| = 0$
$\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}$
$\left| {{\mkern 1mu} \begin{array}{*{20}{c}} 2&{{{\sin }^2}\theta }&{4\sin 3\theta }\\ 2&{1 + {{\sin }^2}\theta }&{4\sin 3\theta }\\ 1&{{{\sin }^2}\theta }&{1 + 4\sin 3\theta } \end{array}{\mkern 1mu} } \right| = 0$
$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}, \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$
$\left| {{\mkern 1mu} \begin{array}{*{20}{c}} 0&{ – 1}&0\\ 1&1&{ – 1}\\ 1&{{{\sin }^2}\theta }&{1 + 4\sin 3\theta } \end{array}{\mkern 1mu} } \right| = 0$
or $4 \sin 3 \theta=-2$
$\sin 3 \theta=-\frac{1}{2}$
$\theta=\frac{7 \pi}{18}$
