106 If $x$ varies in $\left[0, \frac{\pi}{4}\right]$, then the range of positive numbers $a$ for which the equation $\cos x + \sqrt{a} \sin x = \sqrt{a}$ has a solution is
$\qquad$.
$106[1,3+2 \sqrt{2}]$. The original equation can be transformed into
$$
\sqrt{1-\sin ^{2} x}=\sqrt{a}(1-\sin x),
$$

Squaring both sides, we get
$$
(1+a) \sin ^{2} x-2 a \sin x+a-1=0,
$$

Thus, $\sin x=1$ or $\sin x=\frac{a-1}{a+1}$, and since $x \in\left[0, \frac{\pi}{4}\right]$, we have
$$
\sin x \neq 1,
$$

Therefore,
$$
\sin x=\frac{a-1}{a+1},
$$

Hence,
$$
0 \leqslant \frac{a-1}{a+1} \leqslant \frac{\sqrt{2}}{2},
$$

Solving this, we get $1 \leqslant a \leqslant 3+2 \sqrt{2}$.