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

& \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{aligned}$$ Therefore, the equation now becomes as this $$\begin{aligned} & 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{aligned}$$ Now, using the identities of trigonometry to convert sin (3 $$\theta $$ ) and cos (2 $$\theta $$ ) to basic form of sin $$\theta $$ Hence, finally we get the equation as $$\begin{aligned} & 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{aligned}$$ Now, as the equation is in a single function that is sin $$\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{aligned} & 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{aligned}$$ (This is because sin $$\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.