Before we get to your question

$-\ 1 \le sin(x) \le 1\ and\ -\ 1 \le cos(x) \le 1 \implies$

$-\ 2 \le sin(x) + cos(x) \le 2 \implies$

$\sqrt{sin(x) + cos(x)} \in \mathbb R\ only\ if\ 0 \le sin(x) + cos(x) \le 2 \implies$

$0 \le x \le \sqrt{2}\ if\ x \in \mathbb R\ and\ x = \sqrt{sin(x) + cos(x)}.$

But your answer implies that $x > \sqrt{2}.$

So I am confused from the get-go.

Now consider the potential x-intercepts in $[0,\ \sqrt{2}]$ for the function

$f(x) = \sqrt{sin(x) + cos(x)} - x \implies f(0) = 1.$

Moreover, $f'(x) = -\ \dfrac{1}{2\sqrt{sin(x) + cos(x)}} - 1 < 0.$

So what is the maximum number of possible x-intercepts in $[0,\ \sqrt{2}]$