Question 404 of 472HardMCQ
The number of \[\theta \in(0,4 \pi)\] for which the system of linear equations
\[
3(sin3θ)x−y+z=23(cos2θ)x+4y+3z=36x+7y+7z=9
\]
has no solution, is :
Solutions · 1
Official solution
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
\[\therefore\] \[D = \left| {3sin3θ3cos2θ6−147137 } \right| = 0\]
\[ \Rightarrow 3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0\]
\[ \Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0\]
\[ \Rightarrow \sin 3\theta + 2\cos 2\theta - 2 = 0\]
\[ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta + 2(1 - 2{\sin ^2}\theta ) - 2 = 0\]
\[ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta - 4{\sin ^2}\theta = 0\]
\[ \Rightarrow 4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0\]
\[ \Rightarrow \sin \theta (4{\sin ^2}\theta + 4\sin \theta - 3) = 0\]
\[\therefore\] \[\sin \theta = 0\]
\[ \Rightarrow \theta = \pi ,2\pi ,3\pi \] 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.
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