1. **State the problem:** Solve the equation $$\sec^2\theta = 5(\tan\theta - 1)$$ for $$0^\circ \leq \theta \leq 360^\circ$$. 2. **Recall the identity:** $$\sec^2\theta = 1 + \tan^2\theta$$. 3. **Substitute the identity into the equation:** $$1 + \tan^2\theta = 5(\tan\theta - 1)$$ 4. **Expand the right side:** $$1 + \tan^2\theta = 5\tan\theta - 5$$ 5. **Bring all terms to one side to form a quadratic in $$\tan\theta$$:** $$\tan^2\theta - 5\tan\theta + 6 = 0$$ 6. **Factor the quadratic:** $$ (\tan\theta - 2)(\tan\theta - 3) = 0 $$ 7. **Set each factor equal to zero and solve for $$\tan\theta$$:** - $$\tan\theta = 2$$ - $$\tan\theta = 3$$ 8. **Find $$\theta$$ values in $$0^\circ \leq \theta \leq 360^\circ$$ where $$\tan\theta = 2$$:** - $$\theta = \arctan(2) \approx 63.43^\circ$$ - Since tangent is positive in the third quadrant, $$\theta = 180^\circ + 63.43^\circ = 243.43^\circ$$ 9. **Find $$\theta$$ values where $$\tan\theta = 3$$:** - $$\theta = \arctan(3) \approx 71.57^\circ$$ - Tangent positive in third quadrant again, $$\theta = 180^\circ + 71.57^\circ = 251.57^\circ$$ 10. **Final solutions:** $$\theta \approx 63.43^\circ, 243.43^\circ, 71.57^\circ, 251.57^\circ$$ These are the four angles in the given interval satisfying the equation.