Easy as you Like

$sinx + cosx = 1 \rightarrow \sqrt2 (sin(x+\frac{\pi}{4})) \rightarrow sin(x+\frac{\pi}{4}) = \frac {\sqrt2}{2}$

$\rightarrow sin(x+\frac{\pi}{4}) = sin (\frac {\pi}{4})$

$\quad$

Case1:

$x + \frac{\pi}{4} = 2n\pi + \frac{\pi}{4} \rightarrow x = 2n\pi$

Case2:

$x+ \frac{\pi}{4} = 2n\pi + \pi - \frac{\pi}{4} \rightarrow x = (2n+1)\pi + \frac{\pi}{2}$

$\quad$

Note:

Case 2 answer is equal to $n\pi + \frac{\pi}{2}$ .

$\quad$

P.S:This doesn't have anything to do with combinatorics?!

0 and pi/2 along with multiple 2pi revolutions is the general form but there is no way to yield x=0 as a solution.so the correct answer is

2 distinct options only.

If sin(x) + cos(x) = 1, then sin^(x) = 1 - cos(x), and squaring, sin^2(x) = 1 - 2cos(x) + cos^2(x). Adding cos^2(X) to both sides, 1 = 2cos^2(x) - 2cos(x) + 1. Subtracting 1 from each side, transposing 2cos(x) and dividing by 2, we have cos(x) (cos(x) - 1) = 0. So there are 2 choices: (1) cos(x) = 0, whence x = +/- 2n pi, and cos(x) = 0, whence x = +/- (2k+1)*pi/2. Ed Gray