Trigonometrical Equations 1 Question 27
27. Consider the system of linear equations in $x, y, z$
$$ \begin{aligned} (\sin 3 \theta) x-y+z & =0, \\ (\cos 2 \theta) x+4 y+3 z & =0, \\ 2 x+7 y+7 z & =0 \end{aligned} $$
Find the values of $\theta$ for which this system has non-trivial solutions.
(1986, 4M)
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Answer:
Correct Answer: 27. $\pm \frac{\pi}{3}, \pm \frac{2 \pi}{3}$
Solution:
- Since, the given system has non-trivial solution.
$$ \begin{array}{lc} \therefore & \left|\begin{array}{ccc} \sin 3 \theta & -1 & 1 \\ \cos 2 \theta & 4 & 3 \\ 2 & 7 & 7 \end{array}\right|=0 \\ \Rightarrow & \sin 3 \theta(28-21)+1(7 \cos 2 \theta-6) \\ \Rightarrow & +1(7 \cos 2 \theta-8)=0 \\ \Rightarrow & 7 \sin 3 \theta+14 \cos 2 \theta-14=0 \\ \Rightarrow & 3 \sin \theta-4 \sin ^{3} \theta+2\left(1-2 \sin ^{2} \theta\right)-2=0 \end{array} $$
$$ \begin{array}{lc} \Rightarrow & \sin \theta\left(4 \sin ^{2} \theta+4 \sin \theta-3\right)=0 \\ \Rightarrow & \sin \theta=0 \\ \Rightarrow & \theta=n \pi \\ \text { or } & 4 \sin ^{2} \theta+4 \sin \theta-3=0 \\ \Rightarrow & (2 \sin \theta-1)(2 \sin \theta+3)=0 \\ \Rightarrow & \sin \theta=\frac{1}{2} \quad\left[\because \sin \theta=-\frac{3}{2} \text { is not possible }\right] \\ \therefore & \theta=n \pi+(-1)^{n} \frac{\pi}{6} \end{array} $$
$\therefore$ From Eqs. (i) and (ii), we get
$$ \theta=n \pi \text { or } n \pi+(-1)^{n} \frac{\pi}{6} $$
