1. **State the problem:** We need to find all values of $x$ in the interval $0 \leq x \leq 2\pi$ such that $$\cos\frac{11\pi}{31} = \sin x.$$\n\n2. **Use the identity:** Recall that $$\sin x = \cos\left(\frac{\pi}{2} - x\right).$$ So the equation becomes $$\cos\frac{11\pi}{31} = \cos\left(\frac{\pi}{2} - x\right).$$\n\n3. **Solve for angles:** For $\cos A = \cos B$, the solutions are $$A = B + 2k\pi \quad \text{or} \quad A = -B + 2k\pi,$$ where $k$ is any integer.\n\n4. **Apply to our problem:** Let $$A = \frac{11\pi}{31}, \quad B = \frac{\pi}{2} - x.$$ Then,\n$$\frac{11\pi}{31} = \frac{\pi}{2} - x + 2k\pi \quad \Rightarrow \quad x = \frac{\pi}{2} - \frac{11\pi}{31} + 2k\pi,$$\nand\n$$\frac{11\pi}{31} = -\left(\frac{\pi}{2} - x\right) + 2k\pi = -\frac{\pi}{2} + x + 2k\pi \quad \Rightarrow \quad x = \frac{11\pi}{31} + \frac{\pi}{2} + 2k\pi.$$\n\n5. **Simplify expressions:**\n$$x = \frac{\pi}{2} - \frac{11\pi}{31} + 2k\pi = \frac{31\pi}{62} - \frac{22\pi}{62} + 2k\pi = \frac{9\pi}{62} + 2k\pi,$$\n$$x = \frac{11\pi}{31} + \frac{\pi}{2} + 2k\pi = \frac{22\pi}{62} + \frac{31\pi}{62} + 2k\pi = \frac{53\pi}{62} + 2k\pi.$$\n\n6. **Find all $x$ in $[0, 2\pi]$:** Since $2\pi = \frac{124\pi}{62}$, check $k=0$ and $k=1$ for each solution.\n- For $x = \frac{9\pi}{62} + 2k\pi$: \n - $k=0$: $x = \frac{9\pi}{62}$ (valid)\n - $k=1$: $x = \frac{9\pi}{62} + \frac{124\pi}{62} = \frac{133\pi}{62} > 2\pi$ (invalid)\n- For $x = \frac{53\pi}{62} + 2k\pi$: \n - $k=0$: $x = \frac{53\pi}{62}$ (valid)\n - $k=1$: $x = \frac{53\pi}{62} + \frac{124\pi}{62} = \frac{177\pi}{62} > 2\pi$ (invalid)\n\n7. **Final answer:** The solutions in $[0, 2\pi]$ are $$x = \frac{9\pi}{62} \quad \text{and} \quad x = \frac{53\pi}{62}.$$