1. **State the problem:** Solve the equation $$1 + \sin x / \sec x = \cos^3 x / (1 - \sin x)$$ for $x$. 2. **Recall identities:** Recall that $\sec x = \frac{1}{\cos x}$, so $\frac{\sin x}{\sec x} = \sin x \cdot \cos x$. 3. **Rewrite the equation:** Substitute to get $$1 + \sin x \cos x = \frac{\cos^3 x}{1 - \sin x}$$ 4. **Multiply both sides by $1 - \sin x$ to clear the denominator:** $$\left(1 + \sin x \cos x\right)(1 - \sin x) = \cos^3 x$$ 5. **Expand the left side:** $$1 \cdot (1 - \sin x) + \sin x \cos x (1 - \sin x) = \cos^3 x$$ $$1 - \sin x + \sin x \cos x - \sin^2 x \cos x = \cos^3 x$$ 6. **Group terms:** $$1 - \sin x + \sin x \cos x - \sin^2 x \cos x - \cos^3 x = 0$$ 7. **Use Pythagorean identity $\sin^2 x = 1 - \cos^2 x$ in the term $-\sin^2 x \cos x$:** $$- (1 - \cos^2 x) \cos x = - \cos x + \cos^3 x$$ 8. **Substitute back:** $$1 - \sin x + \sin x \cos x - \cos x + \cos^3 x - \cos^3 x = 0$$ 9. **Simplify $\cos^3 x - \cos^3 x = 0$:** $$1 - \sin x + \sin x \cos x - \cos x = 0$$ 10. **Group terms:** $$(1 - \sin x) + (\sin x \cos x - \cos x) = 0$$ 11. **Factor $\cos x$ from the second group:** $$(1 - \sin x) + \cos x (\sin x - 1) = 0$$ 12. **Rewrite $(\sin x - 1) = - (1 - \sin x)$:** $$(1 - \sin x) - \cos x (1 - \sin x) = 0$$ 13. **Factor $(1 - \sin x)$:** $$(1 - \sin x)(1 - \cos x) = 0$$ 14. **Set each factor equal to zero:** - $1 - \sin x = 0 \implies \sin x = 1$ - $1 - \cos x = 0 \implies \cos x = 1$ 15. **Find solutions:** - $\sin x = 1$ at $x = \frac{\pi}{2} + 2k\pi$, $k \in \mathbb{Z}$ - $\cos x = 1$ at $x = 2k\pi$, $k \in \mathbb{Z}$ **Final answer:** $$x = \frac{\pi}{2} + 2k\pi \quad \text{or} \quad x = 2k\pi, \quad k \in \mathbb{Z}$$