1. **Problem:** Solve the trigonometric equation $$9 \sin^2 x + 3 \cos x - 7 = 0$$ for $$0^\circ \leq x \leq 360^\circ$$. 2. **Recall the Pythagorean identity:** $$\sin^2 x + \cos^2 x = 1$$. 3. **Rewrite $$\sin^2 x$$ in terms of $$\cos^2 x$$:** $$\sin^2 x = 1 - \cos^2 x$$. 4. **Substitute into the equation:** $$9(1 - \cos^2 x) + 3 \cos x - 7 = 0$$ 5. **Expand and simplify:** $$9 - 9 \cos^2 x + 3 \cos x - 7 = 0$$ $$-9 \cos^2 x + 3 \cos x + 2 = 0$$ 6. **Multiply both sides by -1 to simplify:** $$9 \cos^2 x - 3 \cos x - 2 = 0$$ 7. **Let $$y = \cos x$$, then solve the quadratic:** $$9 y^2 - 3 y - 2 = 0$$ 8. **Use the quadratic formula:** $$y = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 9 \times (-2)}}{2 \times 9} = \frac{3 \pm \sqrt{9 + 72}}{18} = \frac{3 \pm \sqrt{81}}{18} = \frac{3 \pm 9}{18}$$ 9. **Calculate the roots:** - $$y_1 = \frac{3 + 9}{18} = \frac{12}{18} = \frac{2}{3}$$ - $$y_2 = \frac{3 - 9}{18} = \frac{-6}{18} = -\frac{1}{3}$$ 10. **Find $$x$$ such that $$\cos x = \frac{2}{3}$$:** - $$x = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.19^\circ$$ - Since cosine is positive in Quadrants I and IV, the other solution is: $$360^\circ - 48.19^\circ = 311.81^\circ$$ 11. **Find $$x$$ such that $$\cos x = -\frac{1}{3}$$:** - $$x = \cos^{-1}\left(-\frac{1}{3}\right) \approx 109.47^\circ$$ - Cosine is negative in Quadrants II and III, so the other solution is: $$360^\circ - 109.47^\circ = 250.53^\circ$$ 12. **Final solutions:** $$x \approx 48.19^\circ, 311.81^\circ, 109.47^\circ, 250.53^\circ$$ These are all solutions in the interval $$0^\circ \leq x \leq 360^\circ$$.