1. Rewrite the equation in terms of a single variable

Let $t=\sin x.$ Since $x\in\left(0,\frac\pi2\right)$, we have $0<t<1.$

The equation becomes $\frac{4}{t}+\frac{1}{1-t}=\alpha.$

So we need the minimum value of $f(t)=\frac{4}{t}+\frac{1}{1-t}, \qquad 0<t<1.$

2. Differentiate to find the minimum

$f(t)=4t^{-1}+(1-t)^{-1}$

Differentiate: $f'(t)=-\frac{4}{t^2}+\frac{1}{(1-t)^2}.$

For critical points, set $f'(t)=0$: $-\frac{4}{t^2}+\frac{1}{(1-t)^2}=0$ $\frac{1}{(1-t)^2}=\frac{4}{t^2}$ $t^2=4(1-t)^2.$

Since $0<t<1$, both sides are positive, so we can write $t=2(1-t).$ Thus, $t=2-2t$ $3t=2$ $t=\frac23.$

3. Check that this gives a minimum

As $t\to 0^+$, $f(t)=\frac{4}{t}+\frac{1}{1-t}\to \infty.$

As $t\to 1^-,$ $f(t)=\frac{4}{t}+\frac{1}{1-t}\to \infty.$

So the interior critical point $t=\frac23$ gives the minimum value.

4. Compute the minimum value

$f\left(\frac23\right)=\frac{4}{2/3}+\frac{1}{1-2/3}$ $=4\cdot \frac32+\frac{1}{1/3}$ $=6+3=9.$

Hence the minimum value of $\alpha$ for which the equation has at least one solution is $\boxed{9}.$

5. Comparison with stored answer

Stored correct answer = $9$.

Our derived answer is also $9$, so they agree.