The solutions of the equation1+sin2xsin2xsin2xcos2x1+cos2xcos2x4 sin 2x4 sin 2x1+4 sin 2x=0, (0<x<π) are

R1→R1+R2+R3 ⇒ (2+4sin2x) 1 1 1cos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x=0 sin2x=-12 2x=π+π6, 2π-π6 x=π2+π12, π-π12 =7π12, 11π12

The solutions of the equation1+sin2xsin2xsin2xcos2x1+cos2xcos2x4 sin 2x4 sin 2x1+4 sin 2x=0, (0<x<π) are

R1→R1+R2+R3 ⇒ (2+4sin2x) 1 1 1cos2x1+cos2xcos2x4sin2x4sin2x1+4sin2x=0 sin2x=-12 2x=π+π6, 2π-π6 x=π2+π12, π-π12 =7π12, 11π12

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The solutions of the equation
$|\begin{matrix} 1 + \sin^{2} x & \sin^{2} x & \sin^{2} x \\ \cos^{2} x & 1 + \cos^{2} x & \cos^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0, (0 < x < π) are$

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$\frac{7 π}{12}, \frac{11 π}{12}$

$\frac{5 π}{12}, \frac{7 π}{12}$

$\frac{π}{6}, \frac{5 π}{6}$

$\frac{π}{12}, \frac{π}{6}$

answer is B.

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Detailed Solution

$R_{1} \to R_{1} + R_{2} + R_{3} \Rightarrow (2 + 4 \sin 2 x) |\begin{matrix} 1 & 1 & 1 \\ \cos^{2} x & 1 + \cos^{2} x & \cos^{2} x \\ 4 \sin 2 x & 4 \sin 2 x & 1 + 4 \sin 2 x \end{matrix}| = 0 \sin 2 x = \frac{- 1}{2} 2 x = π + \frac{π}{6}, 2 π - \frac{π}{6} x = \frac{π}{2} + \frac{π}{12}, π - \frac{π}{12} = \frac{7 π}{12}, \frac{11 π}{12}$