JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 4)
Let $$\theta \in \left( {0,{\pi \over 2}} \right)$$. If the system of linear equations
$$(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0$$
$${\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0$$
$${\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0$$
has a non-trivial solution, then the value of $$\theta$$ is :
$$(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0$$
$${\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0$$
$${\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0$$
has a non-trivial solution, then the value of $$\theta$$ is :
$${{4\pi } \over 9}$$
$${{7\pi } \over {18}}$$
$${\pi \over {18}}$$
$${{5\pi } \over {18}}$$
Explanation
$$\left| {\matrix{
{1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\sin 3\,\theta } \cr
{{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\sin 3\,\theta } \cr
{{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr
} } \right| = 0$$
$${C_1} \to {C_1} + {C_2}$$
$$\left| {\matrix{ 2 & {{{\sin }^2}\theta } & {4\sin 3\,\theta } \cr 2 & {1 + {{\sin }^2}\theta } & {4\sin 3\,\theta } \cr 1 & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr } } \right| = 0$$
$${R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3}$$
$$\left| {\matrix{ 0 & { - 1} & 0 \cr 1 & 1 & { - 1} \cr 1 & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr } } \right| = 0$$
or $$4\sin 3\theta = - 2$$
$$\sin 3\theta = - {1 \over 2}$$
$$\theta = {{7\pi } \over {18}}$$
$${C_1} \to {C_1} + {C_2}$$
$$\left| {\matrix{ 2 & {{{\sin }^2}\theta } & {4\sin 3\,\theta } \cr 2 & {1 + {{\sin }^2}\theta } & {4\sin 3\,\theta } \cr 1 & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr } } \right| = 0$$
$${R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3}$$
$$\left| {\matrix{ 0 & { - 1} & 0 \cr 1 & 1 & { - 1} \cr 1 & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \cr } } \right| = 0$$
or $$4\sin 3\theta = - 2$$
$$\sin 3\theta = - {1 \over 2}$$
$$\theta = {{7\pi } \over {18}}$$
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