Solve the equation on the interval \( 0 \leq 0<2 \pi \) \( \sec \frac{3 \theta}{2}=-2 \)

To solve the equation \( \sec \frac{3 \theta}{2} = -2 \), we first rewrite it in terms of cosine: \[ \cos \frac{3 \theta}{2} = -\frac{1}{2}. \] The cosine function equals \(-\frac{1}{2}\) at angles of: \[ \frac{3 \theta}{2} = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad \frac{3 \theta}{2} = \frac{4\pi}{3} + 2k\pi, \] where \( k \) is any integer. Next, we will solve for \( \theta \) in each case. **For the first case:** \[ \frac{3 \theta}{2} = \frac{2\pi}{3} + 2k\pi \Rightarrow 3 \theta = \frac{4\pi}{3} + 4k\pi \Rightarrow \theta = \frac{4\pi}{9} + \frac{4k\pi}{3}. \] **For \( k = 0 \):** \[ \theta = \frac{4\pi}{9}. \] **For \( k = 1 \):** \[ \theta = \frac{4\pi}{9} + \frac{4\pi}{3} = \frac{4\pi}{9} + \frac{12\pi}{9} = \frac{16\pi}{9}, \] which is greater than \( 2\pi \). **For \( k = -1 \):** \[ \theta = \frac{4\pi}{9} - \frac{4\pi}{3} = \frac{4\pi}{9} - \frac{12\pi}{9} = -\frac{8\pi}{9}, \] which is negative. So from the first case, the only valid solution within the interval is \( \theta = \frac{4\pi}{9} \). **For the second case:** \[ \frac{3 \theta}{2} = \frac{4\pi}{3} + 2k\pi \Rightarrow 3 \theta = \frac{8\pi}{3} + 4k\pi \Rightarrow \theta = \frac{8\pi}{9} + \frac{4k\pi}{3}. \] **For \( k = 0 \):** \[ \theta = \frac{8\pi}{9}. \] **For \( k = 1 \):** \[ \theta = \frac{8\pi}{9} + \frac{4\pi}{3} = \frac{8\pi}{9} + \frac{12\pi}{9} = \frac{20\pi}{9}, \] which is greater than \( 2\pi \). **For \( k = -1 \):** \[ \theta = \frac{8\pi}{9} - \frac{4\pi}{3} = \frac{8\pi}{9} - \frac{12\pi}{9} = -\frac{4\pi}{9}, \] which is negative. Thus, the only valid solution from the second case is \( \theta = \frac{8\pi}{9} \). In conclusion, the solutions to the equation \( \sec \frac{3 \theta}{2} = -2 \) in the interval \( 0 \leq \theta < 2\pi \) are: \[ \theta = \frac{4\pi}{9}, \quad \theta = \frac{8\pi}{9}. \]