1. **State the problem:** Solve the equation $$3\cos^3\theta + 3\cos\theta + 4\cos\theta - 4 = 0$$ for $$0^\circ < \theta \leq 270^\circ$$ and determine the number of roots. 2. **Combine like terms:** $$3\cos^3\theta + (3\cos\theta + 4\cos\theta) - 4 = 0$$ $$3\cos^3\theta + 7\cos\theta - 4 = 0$$ 3. **Let $$x = \cos\theta$$, then the equation becomes:** $$3x^3 + 7x - 4 = 0$$ 4. **Solve the cubic equation:** We look for rational roots using the Rational Root Theorem. Possible roots are $$\pm1, \pm\frac{1}{3}, \pm2, \pm\frac{2}{3}, \pm4, \pm\frac{4}{3}$$. 5. **Test $$x=1$$:** $$3(1)^3 + 7(1) - 4 = 3 + 7 - 4 = 6 \neq 0$$ 6. **Test $$x=\frac{1}{3}$$:** $$3\left(\frac{1}{3}\right)^3 + 7\left(\frac{1}{3}\right) - 4 = 3\left(\frac{1}{27}\right) + \frac{7}{3} - 4 = \frac{1}{9} + \frac{7}{3} - 4 = \frac{1}{9} + \frac{21}{9} - \frac{36}{9} = -\frac{14}{9} \neq 0$$ 7. **Test $$x=\frac{4}{3}$$:** $$3\left(\frac{4}{3}\right)^3 + 7\left(\frac{4}{3}\right) - 4 = 3\left(\frac{64}{27}\right) + \frac{28}{3} - 4 = \frac{192}{27} + \frac{252}{27} - \frac{108}{27} = \frac{336}{27} \neq 0$$ 8. **Test $$x=-1$$:** $$3(-1)^3 + 7(-1) - 4 = -3 -7 -4 = -14 \neq 0$$ 9. **Use the cubic formula or numerical methods to find roots:** The cubic has one real root and two complex roots (since the discriminant is negative). The real root is approximately $$x \approx 0.5$$. 10. **Find corresponding $$\theta$$:** Since $$x = \cos\theta$$, $$\cos\theta = 0.5$$. 11. **Find $$\theta$$ in $$0^\circ < \theta \leq 270^\circ$$:** $$\theta = 60^\circ$$ and $$\theta = 300^\circ$$, but $$300^\circ$$ is outside the interval. 12. **Check for other roots:** Since the cubic has only one real root, there is only one $$\theta$$ in the interval. **Final answer:** The equation has **1** root in $$0^\circ < \theta \leq 270^\circ$$. **Answer choice:** A. 1