1. **Problem statement:** Given $a\cos\theta - b\sin\theta = c$ with $a = b = c = 1$, find the value of $\theta$ where $0 \leq \theta \leq 2\pi$. 2. **Formula and rules:** The equation is $\cos\theta - \sin\theta = 1$. 3. **Rewrite the equation:** $$\cos\theta - \sin\theta = 1$$ 4. **Square both sides to use Pythagorean identity:** $$ (\cos\theta - \sin\theta)^2 = 1^2 $$ $$ \cos^2\theta - 2\cos\theta\sin\theta + \sin^2\theta = 1 $$ 5. **Use identity $\cos^2\theta + \sin^2\theta = 1$:** $$ 1 - 2\cos\theta\sin\theta = 1 $$ 6. **Simplify:** $$ -2\cos\theta\sin\theta = 0 $$ $$ \Rightarrow \cos\theta\sin\theta = 0 $$ 7. **Solve for $\theta$:** Either $\cos\theta = 0$ or $\sin\theta = 0$. - If $\cos\theta = 0$, then $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$. - If $\sin\theta = 0$, then $\theta = 0, \pi, 2\pi$. 8. **Check original equation for these values:** - At $\theta = 0$: $\cos0 - \sin0 = 1 - 0 = 1$ (valid) - At $\theta = \pi$: $\cos\pi - \sin\pi = -1 - 0 = -1$ (not valid) - At $\theta = 2\pi$: $\cos2\pi - \sin2\pi = 1 - 0 = 1$ (valid) - At $\theta = \frac{\pi}{2}$: $\cos\frac{\pi}{2} - \sin\frac{\pi}{2} = 0 - 1 = -1$ (not valid) - At $\theta = \frac{3\pi}{2}$: $\cos\frac{3\pi}{2} - \sin\frac{3\pi}{2} = 0 - (-1) = 1$ (valid) 9. **Final answer:** $$ \theta = 0, \frac{3\pi}{2}, 2\pi $$