Use an identity to solve the equation on the interval \( [0,2 \pi) \) \[ \sin 2 x \cos x+\cos 2 x \sin x=-\frac{\sqrt{2}}{2} \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( x=\square \) (Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed. Type your answer in radians. Simplify your answer. Use integers or \( t \) any numbers in the expression.) B. There is no solution.

To solve the equation \(\sin 2x \cos x + \cos 2x \sin x = -\frac{\sqrt{2}}{2}\), we can use the sine addition formula: \[ \sin(2x + x) = \sin(3x) \] Thus, the equation can be rewritten as: \[ \sin(3x) = -\frac{\sqrt{2}}{2} \] Now, we need to find the angles where \(\sin(3x) = -\frac{\sqrt{2}}{2}\). The sine function is negative in the third and fourth quadrants. The reference angle that gives \(\sin = \frac{\sqrt{2}}{2}\) is \(\frac{\pi}{4}\). Therefore, the solutions for \(3x\) can be given by: \[ 3x = \frac{7\pi}{4} + 2k\pi \quad \text{and} \quad 3x = \frac{5\pi}{4} + 2k\pi, \quad k \in \mathbb{Z} \] Dividing by 3 to find \(x\): 1. \(x = \frac{7\pi}{12} + \frac{2k\pi}{3}\) 2. \(x = \frac{5\pi}{12} + \frac{2k\pi}{3}\) Next, we will consider \(k = 0\) since we want solutions in the interval \([0, 2\pi)\). 1. For \(x = \frac{7\pi}{12}\): - \(k = 0\): \(x = \frac{7\pi}{12} \approx 1.83 \text{ (in } [0, 2\pi))\) 2. For \(x = \frac{5\pi}{12}\): - \(k = 0\): \(x = \frac{5\pi}{12} \approx 1.32 \text{ (in } [0, 2\pi))\) Next, consider \(k = 1\): 1. \(x = \frac{7\pi}{12} + \frac{2\pi}{3} = \frac{7\pi}{12} + \frac{8\pi}{12} = \frac{15\pi}{12} = \frac{5\pi}{4}\) (Still in \([0, 2\pi)\)) 2. \(x = \frac{5\pi}{12} + \frac{2\pi}{3} = \frac{5\pi}{12} + \frac{8\pi}{12} = \frac{13\pi}{12}\) (Still in \([0, 2\pi)\)) When \(k = 2\), both values exceed \(2\pi\), as \(\frac{7\pi}{12} + \frac{4\pi}{3} > 2\pi\) and \(\frac{5\pi}{12} + \frac{4\pi}{3} > 2\pi\). Thus, the solutions in the interval \([0, 2\pi)\) are: \[ x = \frac{7\pi}{12}, \frac{5\pi}{12}, \frac{5\pi}{4}, \frac{13\pi}{12} \] Putting them in the answer format: A. \( x = \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{5\pi}{4}, \frac{13\pi}{12} \)