1. **State the problem:** Solve the equation $\sin x + \cos x = 1$ for $0 \leq x \leq 2\pi$. 2. **Use the formula:** Recall the identity $\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$. 3. **Rewrite the equation:** $$\sin x + \cos x = 1 \implies \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = 1$$ 4. **Isolate the sine term:** $$\sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ 5. **Find general solutions for sine:** $$x + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad x + \frac{\pi}{4} = \pi - \frac{\pi}{4} + 2k\pi$$ which simplifies to $$x + \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad x + \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi$$ 6. **Solve for $x$:** $$x = 2k\pi \quad \text{or} \quad x = \frac{3\pi}{4} - \frac{\pi}{4} + 2k\pi = \frac{\pi}{2} + 2k\pi$$ 7. **Find solutions in $[0, 2\pi]$ by substituting integer values of $k$:** - For $k=0$: $x=0$ and $x=\frac{\pi}{2}$ - For $k=1$: $x=2\pi$ and $x=\frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$ (outside interval) 8. **Final solutions:** $$x = 0, \frac{\pi}{2}, 2\pi$$ These are the values of $x$ in $[0, 2\pi]$ satisfying $\sin x + \cos x = 1$.