The solutions of the equation $\left|\begin{array}{ccc}1+\sin ^{2} x & \sin ^{2} x & \sin ^{2} x \\ \cos ^{2} x & 1+\cos ^{2} x & \cos ^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1+4 \sin 2 x\end{array}\right|=0,(0<x<\pi)$, are :

(A) $\frac{7 \pi}{12}, \frac{11 \pi}{12}$

(B) $\frac{\pi}{6}, \frac{5 \pi}{6}$

(C) $\frac{\pi}{12}, \frac{\pi}{6}$

(D) $\frac{5 \pi}{12}, \frac{7 \pi}{12}$

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Determinants

The solutions of the equation

1 + sin^{2} x cos^{2} x 4 sin 2 x sin^{2} x 1 + cos^{2} x 4 sin 2 x sin^{2} x cos^{2} x 1 + 4 sin 2 x = 0, (0 < x < π)

, are :

Answer: (A)

In each row the three entries differ only by the

1

sitting on the diagonal, so subtracting the third column from the first and from the second reduces the determinant to

10 - 1 01 - 1 sin^{2} x cos^{2} x 1 + 4 sin 2 x

Expanding along the first row gives

(1 + 4 sin 2 x + cos^{2} x) + sin^{2} x = 2 + 4 sin 2 x

Setting it to zero gives

sin 2 x = - \frac{1}{2}

. For

0 < x < π

the angle

2 x

runs over

(0, 2 π)

, where

sin 2 x = - \frac{1}{2}

2 x = \frac{7 π}{6}

and

2 x = \frac{11 π}{6}

. The solutions are

x = \frac{7 π}{12}

and

x = \frac{11 π}{12}

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option