The values of $\theta $ lying between $\theta = 0$ and $\theta = \frac{\pi }{2}$ and satisfying the equation $\left| {\begin{array}{*{20}{c}} {1 + {{\cos }^2}\theta }&{{{\sin }^2}\theta }&{4\sin 4\;\theta }\\ {{{\cos }^2}\theta }&{1 + {{\sin }^2}\theta }&{4\sin 4\;\theta }\\ {{{\cos }^2}\theta }&{{{\sin }^2}\theta }&{1 + 4\sin 4\;\theta } \end{array}} \right| = 0$, is

$\frac{{7\;\pi }}{{24}},\frac{{5\;\pi }}{{24}}$

$\frac{{5\,\pi }}{{24}},\frac{\pi }{{24}}$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Medium

1 of 1

Application +4 | -1

The values of

\theta

lying between

\theta = 0

and

\theta = \frac{\pi }{2}

and satisfying the equation

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

, is

View answer and explanation

Correct option: A

\frac{{11\;\pi }}{{24}},\frac{{7\;\pi }}{{24}}

Option D (

\frac{\pi }{{24}},\frac{{11\;\pi }}{{24}}

) is correct because these are the solutions for

\theta

within the given range that satisfy the determinant equation.