1. **State the problem:** Solve the equation $$\sec^2 \theta = 2 \tan \theta + 16$$ for $$\theta$$ in the interval $$[0^\circ, 360^\circ)$$. 2. **Recall the identity:** $$\sec^2 \theta = 1 + \tan^2 \theta$$. 3. **Substitute the identity into the equation:** $$1 + \tan^2 \theta = 2 \tan \theta + 16$$ 4. **Rearrange to form a quadratic in $$\tan \theta$$:** $$\tan^2 \theta - 2 \tan \theta + 1 - 16 = 0$$ $$\tan^2 \theta - 2 \tan \theta - 15 = 0$$ 5. **Use the quadratic formula:** For $$ax^2 + bx + c = 0$$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Here, $$a=1$$, $$b=-2$$, $$c=-15$$. 6. **Calculate the discriminant:** $$b^2 - 4ac = (-2)^2 - 4(1)(-15) = 4 + 60 = 64$$ 7. **Find the roots:** $$\tan \theta = \frac{-(-2) \pm \sqrt{64}}{2(1)} = \frac{2 \pm 8}{2}$$ 8. **Evaluate each root:** - $$\tan \theta = \frac{2 + 8}{2} = \frac{10}{2} = 5$$ - $$\tan \theta = \frac{2 - 8}{2} = \frac{-6}{2} = -3$$ 9. **Find $$\theta$$ for each root in $$[0^\circ, 360^\circ)$$:** - For $$\tan \theta = 5$$, $$\theta = \arctan(5) \approx 78.69^\circ$$ (1st quadrant) - Since tangent is positive in the 3rd quadrant, add $$180^\circ$$: $$78.69^\circ + 180^\circ = 258.69^\circ$$ - For $$\tan \theta = -3$$, $$\theta = \arctan(-3) \approx -71.57^\circ$$ - Adjust to $$[0^\circ, 360^\circ)$$ by adding $$180^\circ$$ (tangent negative in 2nd quadrant): $$180^\circ - 71.57^\circ = 108.43^\circ$$ - Also add $$360^\circ$$ to negative angle for 4th quadrant: $$360^\circ - 71.57^\circ = 288.43^\circ$$ 10. **Final solutions:** $$\theta \approx 78.69^\circ, 258.69^\circ, 108.43^\circ, 288.43^\circ$$ These are the values of $$\theta$$ in $$[0^\circ, 360^\circ)$$ satisfying the original equation.