Question:
The solutions of the equation \(\begin{vmatrix} 1 + \sin^2 x & \sin^2 x & \sin^2 x \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4\sin 2x & 4\sin 2x & 1 + 4\sin 2x \end{vmatrix} = 0\), \((0 \lt x \lt \pi)\), are
The solutions of the equation \(\begin{vmatrix} 1 + \sin^2 x & \sin^2 x & \sin^2 x \\ \cos^2 x & 1 + \cos^2 x & \cos^2 x \\ 4\sin 2x & 4\sin 2x & 1 + 4\sin 2x \end{vmatrix} = 0\), \((0 \lt x \lt \pi)\), are
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Row i of the matrix is the unit row \(e_i\) plus \(a_i(1, 1, 1)\), so the determinant is \(1 + (a_1 + a_2 + a_3) = 2 + 4\sin 2x\).
Updated On: Nov 4, 2026
- \(\frac{\pi}{12}, \frac{\pi}{6}\)
- \(\frac{\pi}{6}, \frac{5\pi}{6}\)
- \(\frac{5\pi}{12}, \frac{7\pi}{12}\)
- \(\frac{7\pi}{12}, \frac{11\pi}{12}\)
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The Correct Option is D
Solution and Explanation
Write the matrix as \(I + a\mathbf{1}^T\) where \(a = (\sin^2 x, \cos^2 x, 4\sin 2x)^T\) and \(\mathbf{1} = (1,1,1)^T\): row i equals \(e_i\) plus \(a_i(1,1,1)\).
For such a matrix, \(\det = 1 + \mathbf{1}^T a = 1 + \sin^2 x + \cos^2 x + 4\sin 2x = 2 + 4\sin 2x\).
Setting it to zero: \(\sin 2x = -\frac{1}{2}\). For \(0 \lt x \lt \pi\), \(2x \in (0, 2\pi)\), so \(2x = \frac{7\pi}{6}\) or \(\frac{11\pi}{6}\), giving \(x = \frac{7\pi}{12}, \frac{11\pi}{12}\).
Answer: (4) \(\frac{7\pi}{12}, \frac{11\pi}{12}\)
For such a matrix, \(\det = 1 + \mathbf{1}^T a = 1 + \sin^2 x + \cos^2 x + 4\sin 2x = 2 + 4\sin 2x\).
Setting it to zero: \(\sin 2x = -\frac{1}{2}\). For \(0 \lt x \lt \pi\), \(2x \in (0, 2\pi)\), so \(2x = \frac{7\pi}{6}\) or \(\frac{11\pi}{6}\), giving \(x = \frac{7\pi}{12}, \frac{11\pi}{12}\).
Answer: (4) \(\frac{7\pi}{12}, \frac{11\pi}{12}\)
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