QJEEM22Ma29dd: JEE Main 2022 Mathematics - Class 12 JEEM22Ma29dd Solution: The number of for which the system of linear equations has no…

Given,

$3(\sin 3\theta )x - y + z = 2$

$3(\cos 2\theta )x + 4y + 3z = 3$

$6x + 7y + 7z = 9$

For no solutions determinant of coefficient will be = 0

$\therefore$ $$D=\left|\begin{array}{ccc}3\sin 3\theta & -1& 1\\ 3\cos 2\theta & 4& 3\\ 6& 7& 7\end{array}\right|=0$$

$\Rightarrow 3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0$

$\Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0$

$\Rightarrow \sin 3\theta + 2\cos 2\theta - 2 = 0$

$\Rightarrow 3\sin \theta - 4{\sin ^3}\theta + 2(1 - 2{\sin ^2}\theta ) - 2 = 0$

$\Rightarrow 3\sin \theta - 4{\sin ^3}\theta - 4{\sin ^2}\theta = 0$

$\Rightarrow 4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0$

$\Rightarrow \sin \theta (4{\sin ^2}\theta + 4\sin \theta - 3) = 0$

$\therefore$ $\sin \theta = 0$

$\Rightarrow \theta = \pi ,2\pi ,3\pi$ when $\theta \in (0,4\pi )$

or,

$4{\sin ^2}\theta + 4\sin \theta - 3 = 0$

$\Rightarrow 4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0$

$\Rightarrow 2\sin \theta (2\sin \theta + 3) - 1(2\sin \theta + 3) = 0$

$\Rightarrow (2\sin \theta - 1)(2\sin \theta + 3) = 0$

$\therefore$ $\sin \theta = {1 \over 2}$

or,

$\sin \theta = - {3 \over 2}$ [not possible as $\sin \in [ - 1,1]$ ]

$\therefore$ $\sin \theta = {1 \over 2}$

$\Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$

$\therefore$ Possible values of $\theta = \pi ,2\pi ,3\pi ,{\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$

$\therefore$ Total 7 values of $\theta$ possible.