1. **Problem b:** Solve the equation $\log_6 x + \log_6 (x + 1) = 1$. 2. **Formula and rules:** Use the logarithm product rule: $\log_a b + \log_a c = \log_a (bc)$. 3. **Apply the rule:** $$\log_6 x + \log_6 (x + 1) = \log_6 [x(x + 1)] = 1$$ 4. **Rewrite the equation:** $$\log_6 [x(x + 1)] = 1$$ means $$x(x + 1) = 6^1 = 6$$ 5. **Form quadratic equation:** $$x^2 + x = 6$$ $$x^2 + x - 6 = 0$$ 6. **Factorize:** $$(x + 3)(x - 2) = 0$$ 7. **Solutions:** $$x = -3 \quad \text{or} \quad x = 2$$ 8. **Check domain:** Since $\log_6 x$ requires $x > 0$, discard $x = -3$. 9. **Final solution for b:** $$x = 2$$ --- 10. **Problem c:** Solve $\sin 2x = \sin x$ for $0^\circ \leq x \leq 360^\circ$. 11. **Use identity:** $\sin 2x = 2 \sin x \cos x$. 12. **Rewrite equation:** $$2 \sin x \cos x = \sin x$$ 13. **Bring all terms to one side:** $$2 \sin x \cos x - \sin x = 0$$ 14. **Factor out $\sin x$:** $$\sin x (2 \cos x - 1) = 0$$ 15. **Set each factor to zero:** - $\sin x = 0$ - $2 \cos x - 1 = 0$ 16. **Solve $\sin x = 0$:** $$x = 0^\circ, 180^\circ, 360^\circ$$ 17. **Solve $2 \cos x - 1 = 0$:** $$\cos x = \frac{1}{2}$$ 18. **Values of $x$ where $\cos x = \frac{1}{2}$ in $[0^\circ, 360^\circ]$:** $$x = 60^\circ, 300^\circ$$ 19. **Final solutions for c:** $$x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ$$ --- 20. **Problem 15:** Express $\sin 3x + \sin 2x$ as a product. 21. **Use sum-to-product formula:** $$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$ 22. **Apply with $A=3x$, $B=2x$:** $$\sin 3x + \sin 2x = 2 \sin \left( \frac{3x + 2x}{2} \right) \cos \left( \frac{3x - 2x}{2} \right) = 2 \sin \left( \frac{5x}{2} \right) \cos \left( \frac{x}{2} \right)$$ 23. **Final expression:** $$\sin 3x + \sin 2x = 2 \sin \left( \frac{5x}{2} \right) \cos \left( \frac{x}{2} \right)$$