1. **State the problem:** Solve the equation $$15\cot^3\left(3\left(\theta - \frac{\pi}{4}\right)\right) - \cot^2\left(3\left(\theta - \frac{\pi}{4}\right)\right) - 2\cot\left(3\left(\theta - \frac{\pi}{4}\right)\right) = 0$$ for $$0 \leq \theta \leq 4\pi$$. 2. **Rewrite the equation:** Let $$x = \cot\left(3\left(\theta - \frac{\pi}{4}\right)\right)$$. Then the equation becomes: $$15x^3 - x^2 - 2x = 0$$ 3. **Factor the equation:** Factor out $$x$$: $$x(15x^2 - x - 2) = 0$$ 4. **Solve each factor:** - For $$x = 0$$: $$\cot\left(3\left(\theta - \frac{\pi}{4}\right)\right) = 0$$ - For $$15x^2 - x - 2 = 0$$, use the quadratic formula: $$x = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 15 \cdot (-2)}}{2 \cdot 15} = \frac{1 \pm \sqrt{1 + 120}}{30} = \frac{1 \pm \sqrt{121}}{30} = \frac{1 \pm 11}{30}$$ So, $$x_1 = \frac{1 + 11}{30} = \frac{12}{30} = \frac{2}{5}$$ $$x_2 = \frac{1 - 11}{30} = \frac{-10}{30} = -\frac{1}{3}$$ 5. **Solve for $$\theta$$ in each case:** **Case 1: $$x=0$$** $$\cot\left(3\left(\theta - \frac{\pi}{4}\right)\right) = 0$$ Recall $$\cot \alpha = 0$$ when $$\alpha = \frac{\pi}{2} + k\pi$$ for integer $$k$$. So, $$3\left(\theta - \frac{\pi}{4}\right) = \frac{\pi}{2} + k\pi$$ Solve for $$\theta$$: $$\theta = \frac{\pi}{4} + \frac{1}{3}\left(\frac{\pi}{2} + k\pi\right) = \frac{\pi}{4} + \frac{\pi}{6} + \frac{k\pi}{3} = \frac{3\pi}{12} + \frac{2\pi}{12} + \frac{4k\pi}{12} = \frac{5\pi}{12} + \frac{4k\pi}{12}$$ Simplify: $$\theta = \frac{5\pi}{12} + \frac{\pi k}{3}$$ Find all $$\theta$$ in $$[0,4\pi]$$ by choosing integer $$k$$ such that $$0 \leq \theta \leq 4\pi$$. **Case 2: $$x = \frac{2}{5}$$** $$\cot\left(3\left(\theta - \frac{\pi}{4}\right)\right) = \frac{2}{5}$$ Recall $$\cot \alpha = \frac{\cos \alpha}{\sin \alpha}$$, so $$3\left(\theta - \frac{\pi}{4}\right) = \cot^{-1}\left(\frac{2}{5}\right) + k\pi$$ Solve for $$\theta$$: $$\theta = \frac{\pi}{4} + \frac{1}{3}\left(\cot^{-1}\left(\frac{2}{5}\right) + k\pi\right)$$ **Case 3: $$x = -\frac{1}{3}$$** $$\cot\left(3\left(\theta - \frac{\pi}{4}\right)\right) = -\frac{1}{3}$$ Similarly, $$3\left(\theta - \frac{\pi}{4}\right) = \cot^{-1}\left(-\frac{1}{3}\right) + k\pi$$ Solve for $$\theta$$: $$\theta = \frac{\pi}{4} + \frac{1}{3}\left(\cot^{-1}\left(-\frac{1}{3}\right) + k\pi\right)$$ 6. **Summary of solutions:** $$\theta = \frac{5\pi}{12} + \frac{\pi k}{3}$$ $$\theta = \frac{\pi}{4} + \frac{1}{3}\left(\cot^{-1}\left(\frac{2}{5}\right) + k\pi\right)$$ $$\theta = \frac{\pi}{4} + \frac{1}{3}\left(\cot^{-1}\left(-\frac{1}{3}\right) + k\pi\right)$$ where $$k$$ is any integer such that $$0 \leq \theta \leq 4\pi$$. 7. **Interpretation:** Use a calculator to approximate $$\cot^{-1}\left(\frac{2}{5}\right)$$ and $$\cot^{-1}\left(-\frac{1}{3}\right)$$ in radians, then find all $$\theta$$ values in the interval. --- **Final answer:** The solutions are all $$\theta$$ in $$[0,4\pi]$$ satisfying the above three formulas with integer $$k$$.