# A value of

Question:

A value of $\theta \in(0, \pi / 3)$, for which

$\left|\begin{array}{ccc}1+\cos ^{2} \theta & \sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & 1+\sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & \sin ^{2} \theta & 1+4 \cos 6 \theta\end{array}\right|=0$, is :

1. $\frac{7 \pi}{24}$

2. $\frac{\pi}{18}$

3. $\frac{\pi}{9}$

4. $\frac{7 \pi}{36}$

Correct Option: , 3

Solution:

$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}$

$\left|\begin{array}{ccc}1 & -1 & 0 \\ \cos ^{2} \theta & 1+\sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & \sin ^{2} \theta & 1+4 \cos 6 \theta\end{array}\right|=0$

$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$

$\left|\begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & -1 \\ \cos ^{2} \theta & \sin ^{2} \theta & 1+4 \cos 6 \theta\end{array}\right|=0$

$\Rightarrow(1+4 \cos 6 \theta)+\sin ^{2} \theta+1\left(\cos ^{2} \theta\right)=0$

$1+2 \cos 6 \theta=0 \Rightarrow \cos 6 \theta=-1 / 2$

$6 \theta=\frac{2 \pi}{3} \Rightarrow \theta=\frac{\pi}{9}$