Radically defined

$\begin{aligned} \sqrt a \sin \theta - 2 \cos \theta & = \sqrt 2 + \sqrt{2-a} & \small \color{#3D99F6} \text{Note that }\sqrt a, \sqrt{2-a} \ge 0 \implies 0 \le a \le 2 \\ \sqrt{a+4} \sin \left(x - \tan^{-1} \frac 2{\sqrt a}\right) & = \sqrt 2 + \sqrt{2-a} \\ \sin \left(x - \tan^{-1} \frac 2{\sqrt a}\right) & = \frac {\sqrt 2 + \sqrt{2-a}}{\sqrt{a+4}} \end{aligned}$

We note that, for $0 \le a \le 2$ , $\dfrac 1{\sqrt 3} \le \dfrac {\sqrt 2 + \sqrt{2-a}}{\sqrt{a+4}} \le \sqrt 2$ . But $\sin \left(x - \tan^{-1} \frac 2{\sqrt a}\right) \le 1$ , then

$\begin{aligned} \frac {\sqrt 2 + \sqrt{2-a}}{\sqrt{a+4}} & \le 1 \\ \sqrt 2 + \sqrt{2-a} & \le \sqrt{a+4} & \small \color{#3D99F6} \text{Squaring both sides} \\ 2 + 2\sqrt{4-2a} + 2 - a & \le a + 4 & \small \color{#3D99F6} \text{Rearranging} \\ \sqrt{4-2a} & \le a & \small \color{#3D99F6} \text{Squaring both sides again} \\ 4-2a & \le a^2 & \small \color{#3D99F6} \text{Rearranging} \\ a^2 + 2a - 4 & \ge 0 \end{aligned}$

$\begin{aligned} \implies a & \ge \frac {-2 + \sqrt{4+16}}2 = \sqrt 5 -1 & \small \color{#3D99F6} \text{Note that }a > 0 \end{aligned}$

Therefore, $\boxed{\sqrt 5-1 \le a \le 2}$ .