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

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-

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.