Question

Find the number of solutions of the equation
Det \[\left( \begin{matrix}
   \sin 3\theta & -1 & 1 \\
   \cos 2\theta & 4 & 3 \\
   2 & 7 & 7 \\
\end{matrix} \right)=0\] in \[\left[ 0,2\pi \right]\] is \[\]
A.2
B.3
C.4
D.5

Answer 1

The determinant of a 3x3 matrix
     \[\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)\]
 Can be evaluated as

DETERMINANT= \[{{a}_{11}}[\left( {{a}_{22}}{{a}_{33}} \right)-\left( {{a}_{23}}{{a}_{32}} \right)]-{{a}_{12}}\left[ ({{a}_{33}}{{a}_{21}})-({{a}_{23}}{{a}_{31}}) \right]+{{a}_{13}}\left[ ({{a}_{21}}{{a}_{32}})-({{a}_{22}}{{a}_{31}}) \right]\]

We may also use:-
\[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta \]
And
\[\cos 2\theta =1-2{{\sin }^{2}}\theta \]

Complete step-by-step answer:

We know that one can easily evaluate the determinant of a 3x3 matrix.

Now, as per the question the determinant of the question would become
\[\begin{align}
  & \sin 3\theta \cdot (4\times 7-7\times 3)-\cos 2\theta \cdot (-1\times 7-7\times 1)+2\cdot (-1\times 3-4\times 1) \\
 & \sin 3\theta \cdot (28-21)-\cos 2\theta \cdot (-14)+2\times (-7) \\
 & 7\sin 3\theta +14\cos 2\theta -14 \\
\end{align}\]

Therefore, the equation now becomes as this
\[\begin{align}
  & 7\sin 3\theta +14\cos 2\theta -14=0 \\
 & 7(\sin 3\theta +2\cos 2\theta -2)=0 \\
 & \sin 3\theta +2\cos 2\theta -2=0 \\
\end{align}\]

Now, using the identities of trigonometry to convert sin (3 \[\theta \] ) and cos (2 \[\theta \] ) to basic form of sin \[\theta \]

Hence, finally we get the equation as
\[\begin{align}
  & 3\sin \theta -4{{\sin }^{3}}\theta +2(1-{{\sin }^{2}}\theta )-2=0 \\
 & 3\sin \theta -4{{\sin }^{3}}\theta +2-2{{\sin }^{2}}\theta -2=0 \\
 & 3\sin \theta -4{{\sin }^{3}}\theta -2{{\sin }^{2}}\theta =0 \\
\end{align}\]
Now, as the equation is in a single function that is sin \[\theta \] and now we can evaluate further to get the solutions.
\[\sin \theta (3-4{{\sin }^{2}}\theta -2\sin \theta )=0\] (By just taking sin \[\theta \] common from each term)

Now, we have 2 equations,
One is:- \[\sin \theta =0\]
And the other is:- \[4{{\sin }^{2}}\theta +2\sin \theta -3=0\]
 On solving the first one, we get
\[\theta =n\pi \] (Where n is equal to any integer)
And from the second one, we get:-

Taking sin \[\theta \] as t, we can then solve the quadratic equation thus formed
\[\begin{align}
  & 4{{t}^{2}}+2t-3=0 \\
 & t=\dfrac{\pm \sqrt{13}-1}{4} \\
 & t=\dfrac{\pm 3.6-1}{4} \\
 & t=-1.15,0.65 \\
 & \therefore \sin \theta =0.65 \\
\end{align}\]
(This is because sin \[\theta \] cannot assume value less than -1 and more than 1)

Therefore,
\[\theta =n\dfrac{\pi }{\sqrt{2}}\] (Where n is all integers)

Hence, in the given domain, the number of solutions is 5.
\[\left( \dfrac{\pi }{\sqrt{2}},\sqrt{2}\pi ,0,\pi ,2\pi \right)\]\[\]

NOTE: -
We can see here that a lot of concepts are being used here.
The students can make a mistake when checking the domain of sin function and also what domain is given to check in the question.