Question

The values of $\theta $ lying between $0$ and $\dfrac{\pi }{2}$ and satisfying the equation
$\left| {\begin{array}{*{20}{c}}
  {1 + {{\sin }^2}\theta }&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  {{{\sin }^2}\theta }&{1 + {{\cos }^2}\theta }&{4\sin 4\theta } \\
  {{{\sin }^2}\theta }&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }
\end{array}} \right| = 0$
Are
A.$\dfrac{{7\pi }}{{24}}$
B.$\dfrac{{5\pi }}{{24}}$
C.$\dfrac{{11\pi }}{{24}}$
D.$\dfrac{\pi }{{24}}$

Answer 1

Hint: The equation is given in determinant form. We will solve the determinant first, using column and row operations to convert the elements of determinant in simpler form. We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$ so we can perform operation${{\text{C}}_1} \to {{\text{C}}_{\text{1}}}{\text{ + }}{{\text{C}}_2}$. And similarly we can perform row operations to make the determinant easier to solve. Then expand the determinant and solve for$\theta $.

Complete step-by-step answer:
Here the equation is given in the form-
$ \Rightarrow $ $\left| {\begin{array}{*{20}{c}}
  {1 + {{\sin }^2}\theta }&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  {{{\sin }^2}\theta }&{1 + {{\cos }^2}\theta }&{4\sin 4\theta } \\
  {{{\sin }^2}\theta }&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }
\end{array}} \right| = 0$
Here the value of $\theta $ lies between $0$ and$\dfrac{\pi }{2}$. We have to find the value of $\theta $.
Now first we will perform the operation ${{\text{C}}_1} \to {{\text{C}}_{\text{1}}}{\text{ + }}{{\text{C}}_2}$
Then we get-
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + {{\sin }^2}\theta + {{\cos }^2}\theta }&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  {{{\sin }^2}\theta + 1 + {{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{4\sin 4\theta } \\
  {{{\sin }^2}\theta + {{\cos }^2}\theta }&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }
\end{array}} \right| = 0$
Now we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
So on applying this, we get-
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {1 + 1}&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  {1 + 1}&{1 + {{\cos }^2}\theta }&{4\sin 4\theta } \\
  1&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }
\end{array}} \right| = 0$
On simplifying, we get-
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  2&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  2&{1 + {{\cos }^2}\theta }&{4\sin 4\theta } \\
  1&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }
\end{array}} \right| = 0$
Now we can see that the terms on second row match with terms of first row so on performing operation, ${R_2} \to {R_2} - {R_1}$ , we get-
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
  2&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  {2 - 2}&{1 + {{\cos }^2}\theta - {{\cos }^2}\theta }&{4\sin 4\theta - 4\sin 4\theta } \\
  1&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }
\end{array}} \right| = 0$
On simplifying, we get-
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  2&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  0&1&0 \\
  1&{{{\cos }^2}\theta }&{1 + 4\sin 4\theta }
\end{array}} \right| = 0\]
Now performing operation, ${R_3} \to {R_3} - {R_1}$ , we get-
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  2&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  2&1&0 \\
  {1 - 2}&{{{\cos }^2}\theta - {{\cos }^2}\theta }&{1 + 4\sin 4\theta - 4\sin 4\theta }
\end{array}} \right| = 0\]
On simplifying, we get-
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  2&{{{\cos }^2}\theta }&{4\sin 4\theta } \\
  0&1&0 \\
  { - 1}&0&1
\end{array}} \right| = 0\]
Now on expanding the determinant, we get-
\[ \Rightarrow 2\left| {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right| - {\cos ^2}\theta \left| {\begin{array}{*{20}{c}}
  0&0 \\
  { - 1}&1
\end{array}} \right| + 4\sin 4\theta \left| {\begin{array}{*{20}{c}}
  0&1 \\
  { - 1}&0
\end{array}} \right| = 0\]
On solving the above, we get-
\[ \Rightarrow 2\left( {1 - 0} \right) - {\cos ^2}\theta \left( {0 - 0} \right) + 4\sin 4\theta \left( {0 - \left( { - 1} \right)} \right) = 0\]
On simplifying, we get-
\[ \Rightarrow 2 + 4\sin 4\theta = 0\]
On taking $2$ common we get-
\[ \Rightarrow 2\left( {1 + 2\sin 4\theta } \right) = 0\]
On simplifying, we get-
\[ \Rightarrow 1 + 2\sin 4\theta = 0\]
On rearranging, we get-
\[ \Rightarrow 2\sin 4\theta = - 1\]
On simplifying, we get-
\[ \Rightarrow \sin 4\theta = \dfrac{{ - 1}}{2}\]
Here the value of $\sin \theta $ is negative which is only possible in $3rd$ and $4th$ quadrant. And we know that $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$ .Then we can write-
For third quadrant-
$ \Rightarrow \sin 4\theta = \sin \left( {\pi + \dfrac{\pi }{6}} \right)$
On simplifying, we get-
$ \Rightarrow 4\theta = \pi + \dfrac{\pi }{6},$
On taking LCM on the right side, we get-
$ \Rightarrow 4\theta = \dfrac{{7\pi }}{6} $
On transferring $4$ on right side, we get-
$ \Rightarrow \theta = \dfrac{{7\pi }}{{4 \times 6}} = \dfrac{{7\pi }}{{24}}$
For fourth quadrant-
$ \Rightarrow \sin 4\theta = \sin \left( {2\pi - \dfrac{\pi }{6}} \right)$
On solving, we get-
$ \Rightarrow 4\theta = 2\pi - \dfrac{\pi }{6}$
On taking LCM on the right side, we get-
$ \Rightarrow 4\theta = \dfrac{{11\pi }}{6}$
On transferring $4$ on right side, we get-
$ \Rightarrow \theta = \dfrac{{11\pi }}{{4 \times 6}} = \dfrac{{11\pi }}{{24}}$
Hence the correct answers are option A and option C.

Note: Here we can also expand the determinant this way-
\[ \Rightarrow 2\left| {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}
  {{{\cos }^2}\theta }&{4\sin 4\theta } \\
  0&1
\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}
  {{{\cos }^2}\theta }&{4\sin 4\theta } \\
  1&0
\end{array}} \right| = 0\]
On solving this, we get-
\[ \Rightarrow 2\left( {1 - 0} \right) - 0 - 1\left( {0 - 4\sin 4\theta } \right) = 0\]
On simplifying, we get-
\[ \Rightarrow 2 + 4\sin 4\theta = 0\]
So this is also right as solving the solution in the same pattern will give us the same answer.