The solution of sin x + 1 = cos x, such that 0 ≤ x < 2π, is:
Step-by-step Solution:
\[ \sin x + 1 = \cos x \] \[ \cos x - \sin x = 1 \] \[ \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} \] \[ \cos \left(x + \frac{\pi}{4}\right) = \cos \frac{\pi}{4} \] \[ \left(x + \frac{\pi}{4}\right) = 2n\pi \pm \frac{\pi}{4} \] Putting \( n = 0 \) and \( 1 \), \( x = 0 \) and \( \frac{3\pi}{2} \).