1. **State the problem:** Solve the differential equation $$\frac{dy}{dx} = e^x \csc(2y) \csc(y)$$ with initial condition $$y(0) = \frac{\pi}{6}$$. 2. **Rewrite the equation:** Separate variables to isolate $$y$$ terms on one side and $$x$$ terms on the other: $$\frac{dy}{dx} = e^x \csc(2y) \csc(y) \implies \frac{dy}{\csc(2y) \csc(y)} = e^x dx$$ 3. **Simplify the left side:** Recall that $$\csc(\theta) = \frac{1}{\sin(\theta)}$$, so $$\frac{1}{\csc(2y) \csc(y)} = \sin(2y) \sin(y)$$. Thus, $$\sin(2y) \sin(y) dy = e^x dx$$ 4. **Use the double-angle identity:** $$\sin(2y) = 2 \sin(y) \cos(y)$$, so $$\sin(2y) \sin(y) = 2 \sin^2(y) \cos(y)$$. Therefore, $$2 \sin^2(y) \cos(y) dy = e^x dx$$ 5. **Integrate both sides:** $$\int 2 \sin^2(y) \cos(y) dy = \int e^x dx$$ 6. **Substitute for integration:** Let $$u = \sin(y)$$, then $$du = \cos(y) dy$$. Rewrite the left integral: $$\int 2 u^2 du = 2 \int u^2 du = 2 \cdot \frac{u^3}{3} = \frac{2}{3} u^3 + C = \frac{2}{3} \sin^3(y) + C$$ 7. **Integrate the right side:** $$\int e^x dx = e^x + C$$ 8. **Combine results:** $$\frac{2}{3} \sin^3(y) = e^x + C$$ 9. **Apply initial condition:** When $$x=0$$, $$y=\frac{\pi}{6}$$. Calculate $$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$, so $$\frac{2}{3} \left(\frac{1}{2}\right)^3 = \frac{2}{3} \cdot \frac{1}{8} = \frac{1}{12}$$. Thus, $$\frac{1}{12} = e^0 + C = 1 + C \implies C = \frac{1}{12} - 1 = -\frac{11}{12}$$ 10. **Write the implicit solution:** $$\frac{2}{3} \sin^3(y) = e^x - \frac{11}{12}$$ 11. **Solve for $$y$$ explicitly:** $$\sin^3(y) = \frac{3}{2} \left(e^x - \frac{11}{12}\right)$$ $$\sin(y) = \sqrt[3]{\frac{3}{2} \left(e^x - \frac{11}{12}\right)}$$ $$y = \arcsin\left(\sqrt[3]{\frac{3}{2} \left(e^x - \frac{11}{12}\right)}\right)$$ **Final answer:** $$y = \arcsin\left(\sqrt[3]{\frac{3}{2} \left(e^x - \frac{11}{12}\right)}\right)$$