Question

If$f\left(\theta \right)$= $\begin{array}{ccc}1+{\sin }^2\theta & {\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & 1+{\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & {\cos }^2\theta & 1+4\sin 4\theta \end{array}$

Then number of solution of $f\left(\theta \right)=0$ in $[0,\frac{\pi }{2}]$ is

Answer 1

Let $f\left(\theta \right)=\left|\begin{array}{ccc}1+{\sin }^2\theta & {\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & 1+{\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & {\cos }^2\theta & 1+4\sin \theta \end{array}\right|$

[$R_2^{\prime }=R_2-R_1$ and $R_3^{\prime }=R_3-R_1$]

$=\left|\begin{array}{ccc}1+{\sin }^2\theta & {\cos }^2\theta & 4\sin 4\theta \\ -1& 1& 0\\ -1& 0& 1\end{array}\right|$

$=4\sin 4\theta +(1+{\sin }^2\theta +{\cos }^2\theta )$

$=4\sin 4\theta +2$.

Now for $f\left(\theta \right)=0$ we have

$\sin 4\theta =-\frac{1}{2}$

or, $\sin 4\theta =\sin (-\frac{\pi }{6})$

or, $4\theta =n\pi +(-1{)}^{n+1}\frac{\pi }{6}$ [Where $n\in Z$(Set of integers)]

For $n=1$, $4\theta =\pi +\frac{\pi }{6}={210}^o$

$\theta ={52.5}^o$ which $\in [0,\frac{\pi }{2}]$

For $n=2$, $4\theta =2\pi -\frac{\pi }{6}={330}^o$

$\theta ={82.5}^o$ which also $\in [0,\frac{\pi }{2}]$.

For other values of $n$, $\theta$$\notin [0,\frac{\pi }{2}]$.

So, $f\left(\theta \right)=0$ admits two solutions.