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# The number of all possible values of $\theta$, where $0 < \theta < \pi$ , for which the system of equations $\left( y+z \right)\cos 3\theta =\left( xyz \right)\sin 3\theta$  $x\sin 3\theta =\dfrac{2\cos 3\theta }{y}+\dfrac{2\sin 3\theta }{z}$  $\left( xyz \right)\sin 3\theta =\left( y+2z \right)\cos 3\theta +y\sin 3\theta$ Have a solution $\left( x,y,z \right)$ with $yz\ne 0$ , is(a) 0(b) 1(c) 2(d) 3

Last updated date: 20th Jun 2024
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Answer
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Hint: We start solving the problem by simplifying the given system of linear equations by making the necessary calculations. We then solve two of the given three equations to get the relation between the $\theta$ and other variables. We then make use of the result that if $\sin \alpha =\cos \alpha$ , then $\alpha =n\pi +\dfrac{\pi }{4}$ , $n\in Z$ to get the general solution for $\theta$. We then substitute different values for n to get the number of values of $\theta$ that lies in the given interval to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the number of all possible values of $\theta$, where $0 < \theta < \pi$ , for the given system of equations
$\left( y+z \right)\cos 3\theta =\left( xyz \right)\sin 3\theta$
$x\sin 3\theta =\dfrac{2\cos 3\theta }{y}+\dfrac{2\sin 3\theta }{z}$
$\left( xyz \right)\sin 3\theta =\left( y+2z \right)\cos 3\theta +y\sin 3\theta$ , to have a solution $\left( x,y,z \right)$ with $yz\ne 0$ .
Let us simplify the equation $x\sin 3\theta =\dfrac{2\cos 3\theta }{y}+\dfrac{2\sin 3\theta }{z}$ .
$\Rightarrow x\sin 3\theta =\dfrac{2z\cos 3\theta +2y\sin 3\theta }{yz}$ .
$\Rightarrow xyz\sin 3\theta =2z\cos 3\theta +2y\sin 3\theta$ .
Now, let us rewrite the given system of linear equations:
$\Rightarrow y\cos 3\theta +z\cos 3\theta =\left( xyz \right)\sin 3\theta$ ---(1).
$\Rightarrow xyz\sin 3\theta =2z\cos 3\theta +2y\sin 3\theta$ ---(2).
$\Rightarrow \left( xyz \right)\sin 3\theta =y\cos 3\theta +2z\cos 3\theta +y\sin 3\theta$ ---(3).
Let us solve equations (2) and (3).
$\Rightarrow 2z\cos 3\theta +2y\sin 3\theta =y\cos 3\theta +2z\cos 3\theta +y\sin 3\theta$.
$\Rightarrow y\sin 3\theta -y\cos 3\theta =0$.
$\Rightarrow y\left( \sin 3\theta -\cos 3\theta \right)=0$.
We have given that $y\ne 0$ , so we get $\sin 3\theta -\cos 3\theta =0$.
$\Rightarrow \sin 3\theta =\cos 3\theta$ .
We know that if $\sin \alpha =\cos \alpha$ , then $\alpha =n\pi +\dfrac{\pi }{4}$ , $n\in Z$ .
$\Rightarrow 3\theta =n\pi +\dfrac{\pi }{4}$ .
$\Rightarrow \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{12}$ , $n\in Z$ .
Let us substitute $n=0$ , then we get $\theta =\dfrac{\pi }{12}$ .
Now, let us substitute $n=1$ , then we get $\theta =\dfrac{\pi }{3}+\dfrac{\pi }{12}=\dfrac{5\pi }{12}$.
Now, let us substitute $n=2$ , then we get $\theta =\dfrac{2\pi }{3}+\dfrac{\pi }{12}=\dfrac{9\pi }{12}=\dfrac{3\pi }{4}$ .
Since the value of $\theta$ lies in $\left( 0,\pi \right)$ , other values of n will not satisfy.
So, we have found three possible values of $\theta$ .
$\therefore$ The correct option for the given problem is (d).

Note:
We can also solve the given system of linear equations by first writing the coefficient matrix and then equating its determinant to zero which gives us a similar result. We should not make calculation mistakes while solving this problem. We should not forget to find particular solutions after finding the general solution for $\theta$, which is a common mistake done by students. Similarly, we can expect problems to find whether the given system of linear equations has unique or infinite solutions.