1. **State the problem:** Given the equation $\sec x - \tan x = 1$, find the value(s) of $x$. 2. **Recall trigonometric identities:** We know that $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$. 3. **Rewrite the equation using these identities:** $$\frac{1}{\cos x} - \frac{\sin x}{\cos x} = 1$$ 4. **Combine the terms on the left-hand side:** $$\frac{1 - \sin x}{\cos x} = 1$$ 5. **Multiply both sides by $\cos x$ (assuming $\cos x \neq 0$):** $$1 - \sin x = \cos x$$ 6. **Rearrange the equation:** $$1 = \sin x + \cos x$$ 7. **Square both sides to use Pythagorean identity:** $$1^2 = (\sin x + \cos x)^2$$ $$1 = \sin^2 x + 2 \sin x \cos x + \cos^2 x$$ 8. **Use the identity $\sin^2 x + \cos^2 x = 1$:** $$1 = 1 + 2 \sin x \cos x$$ 9. **Simplify:** $$0 = 2 \sin x \cos x$$ 10. **Divide both sides by 2:** $$0 = \sin x \cos x$$ 11. **Set each factor equal to zero:** - $\sin x = 0$ - $\cos x = 0$ 12. **Find solutions for $\sin x = 0$:** $$x = n\pi, \quad n \in \mathbb{Z}$$ 13. **Find solutions for $\cos x = 0$:** $$x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$$ 14. **Check which solutions satisfy the original equation $\sec x - \tan x = 1$:** - For $x = n\pi$, $\sec x = \pm 1$, $\tan x = 0$, so $\sec x - \tan x = \pm 1$. Only when $\sec x = 1$ (i.e., $x = 2n\pi$) does the equation hold. - For $x = \frac{\pi}{2} + n\pi$, $\cos x = 0$, so $\sec x$ is undefined, discard these. 15. **Final solution:** $$x = 2n\pi, \quad n \in \mathbb{Z}$$