1. **State the problem:** We want to find the number of solutions to the equation $$|2 + 3 \sin 2x| = 1$$. 2. **Understand the absolute value equation:** The equation $$|A| = B$$ means $$A = B$$ or $$A = -B$$, provided $$B \geq 0$$. 3. **Apply this to our problem:** We have $$|2 + 3 \sin 2x| = 1 \implies 2 + 3 \sin 2x = 1 \text{ or } 2 + 3 \sin 2x = -1.$$ 4. **Solve each case separately:** **Case 1:** $$2 + 3 \sin 2x = 1$$ $$3 \sin 2x = 1 - 2 = -1$$ $$\sin 2x = -\frac{1}{3}$$ **Case 2:** $$2 + 3 \sin 2x = -1$$ $$3 \sin 2x = -1 - 2 = -3$$ $$\sin 2x = -1$$ 5. **Find solutions for each case:** - For $$\sin 2x = -\frac{1}{3}$$, since sine ranges between -1 and 1, this is valid. - For $$\sin 2x = -1$$, this is also valid. 6. **General solutions for sine equations:** - For $$\sin \theta = a$$, solutions are $$\theta = \arcsin(a) + 2k\pi$$ or $$\theta = \pi - \arcsin(a) + 2k\pi$$ for any integer $$k$$. 7. **Apply to our variable:** Let $$\theta = 2x$$. - For $$\sin 2x = -\frac{1}{3}$$, $$2x = \arcsin\left(-\frac{1}{3}\right) + 2k\pi$$ or $$2x = \pi - \arcsin\left(-\frac{1}{3}\right) + 2k\pi$$. - For $$\sin 2x = -1$$, $$2x = \frac{3\pi}{2} + 2k\pi$$ (since sine equals -1 at $$\frac{3\pi}{2}$$ plus full rotations). 8. **Count solutions in one period:** - The function $$\sin 2x$$ has period $$\pi$$. - Within one period $$x \in [0, \pi)$$, the equation $$\sin 2x = -\frac{1}{3}$$ has 2 solutions. - The equation $$\sin 2x = -1$$ has 1 solution per period. 9. **Total solutions per period:** $$2 + 1 = 3$$ solutions. 10. **Number of solutions depends on the domain:** - If no domain is specified, there are infinitely many solutions because sine is periodic. - If the domain is restricted, count solutions accordingly. **Final answer:** The equation $$|2 + 3 \sin 2x| = 1$$ has infinitely many solutions over all real numbers, with 3 solutions per period of length $$\pi$$ in $$x$$.