1. **State the problem:** Solve the equation $$\sqrt{3} \csc\left(2\theta + \frac{3\pi}{4}\right) = 2$$ for $$-\pi < \theta < \pi$$, giving answers in terms of $$\pi$$. 2. **Rewrite the equation:** Recall that $$\csc x = \frac{1}{\sin x}$$, so the equation becomes: $$\sqrt{3} \cdot \frac{1}{\sin\left(2\theta + \frac{3\pi}{4}\right)} = 2$$ 3. **Isolate the sine term:** $$\frac{\sqrt{3}}{\sin\left(2\theta + \frac{3\pi}{4}\right)} = 2 \implies \sin\left(2\theta + \frac{3\pi}{4}\right) = \frac{\sqrt{3}}{2}$$ 4. **Recall sine values:** $$\sin x = \frac{\sqrt{3}}{2}$$ at $$x = \frac{\pi}{3} + 2k\pi$$ and $$x = \frac{2\pi}{3} + 2k\pi$$ for any integer $$k$$. 5. **Set up equations:** $$2\theta + \frac{3\pi}{4} = \frac{\pi}{3} + 2k\pi$$ or $$2\theta + \frac{3\pi}{4} = \frac{2\pi}{3} + 2k\pi$$ 6. **Solve for $$\theta$$:** For the first: $$2\theta = \frac{\pi}{3} - \frac{3\pi}{4} + 2k\pi = \frac{4\pi}{12} - \frac{9\pi}{12} + 2k\pi = -\frac{5\pi}{12} + 2k\pi$$ $$\theta = -\frac{5\pi}{24} + k\pi$$ For the second: $$2\theta = \frac{2\pi}{3} - \frac{3\pi}{4} + 2k\pi = \frac{8\pi}{12} - \frac{9\pi}{12} + 2k\pi = -\frac{\pi}{12} + 2k\pi$$ $$\theta = -\frac{\pi}{24} + k\pi$$ 7. **Find all solutions in $$-\pi < \theta < \pi$$:** For $$\theta = -\frac{5\pi}{24} + k\pi$$: - If $$k = -1$$, $$\theta = -\frac{5\pi}{24} - \pi = -\frac{5\pi}{24} - \frac{24\pi}{24} = -\frac{29\pi}{24} < -\pi$$ (discard) - If $$k = 0$$, $$\theta = -\frac{5\pi}{24}$$ (valid) - If $$k = 1$$, $$\theta = -\frac{5\pi}{24} + \pi = -\frac{5\pi}{24} + \frac{24\pi}{24} = \frac{19\pi}{24}$$ (valid) For $$\theta = -\frac{\pi}{24} + k\pi$$: - If $$k = -1$$, $$\theta = -\frac{\pi}{24} - \pi = -\frac{\pi}{24} - \frac{24\pi}{24} = -\frac{25\pi}{24} < -\pi$$ (discard) - If $$k = 0$$, $$\theta = -\frac{\pi}{24}$$ (valid) - If $$k = 1$$, $$\theta = -\frac{\pi}{24} + \pi = -\frac{\pi}{24} + \frac{24\pi}{24} = \frac{23\pi}{24}$$ (valid) 8. **Final solutions:** $$\theta = -\frac{5\pi}{24}, \frac{19\pi}{24}, -\frac{\pi}{24}, \frac{23\pi}{24}$$ These are all solutions in the interval $$-\pi < \theta < \pi$$. **Answer:** $$\boxed{\theta = -\frac{5\pi}{24}, \frac{19\pi}{24}, -\frac{\pi}{24}, \frac{23\pi}{24}}$$