minimize of

Let $\sin^2 x=a$ ( $0\leq a\leq 1$ )

Then the given expression is equal to

$\sqrt {a^2+2^2}+\sqrt {(a-1)^2+2^2}$ .

This is actually the sum of two distances :

(i) The distance between points $(0,a)$ and $(2,0)$ and

(ii) The distance between points $(2,0)$ and $(4,a-1)$ .

Obviously, the minimum sum will be the straight line distance between points $(0,a)$ and $(4,a-1)$ , when the three points $(0,a),(2,0),(4,a-1)$ are collinear. Then $a=1-a\implies a=\dfrac {1}{2}$ , and the minimum distance will be

$\sqrt {(\frac {1}{2})^2+4}+\sqrt {(\frac{1}{2}-1)^2+4}=\sqrt {17}\approx \boxed {4.123}$ .