1. We need the minimum value of

$2^{\sin x}+2^{\cos x}.$

2. Let

$a=\sin x,\quad b=\cos x.$ Then

$-1\le a,b\le 1, \qquad a^2+b^2=1.$

So we must minimize

$f(a,b)=2^a+2^b$ subject to

$a^2+b^2=1.$

3. Since the function is symmetric in $a$ and $b$, the minimum should occur when one of them is as small as possible and the other is adjusted by the constraint.

Let us check boundary-type points on the unit circle:

• If $\sin x=-1$, then $\cos x=0$, so $2^{\sin x}+2^{\cos x}=2^{-1}+2^0=\frac12+1=\frac32.$

• If $\cos x=-1$, then $\sin x=0$, so again $2^{\sin x}+2^{\cos x}=1+\frac12=\frac32.$

• If $\sin x=\cos x=\frac{1}{\sqrt2}$, then $2^{\sin x}+2^{\cos x}=2\cdot 2^{1/\sqrt2}=2^{1+1/\sqrt2},$ which is clearly larger.

• If $\sin x=\cos x=-\frac{1}{\sqrt2}$, then $2^{\sin x}+2^{\cos x}=2\cdot 2^{-1/\sqrt2}=2^{1-1/\sqrt2}.$

Now compare

$2^{1-1/\sqrt2} \quad \text{and} \quad \frac32.$

Since $1-\frac1{\sqrt2}\approx 0.2929$,

$2^{1-1/\sqrt2}\approx 2^{0.2929}\approx 1.225<1.5.$

So this is smaller.

4. We now justify that this is indeed the minimum.

Using the convexity of $2^t$ (since $2^t=e^{t\ln2}$ is convex), for fixed $a^2+b^2=1$, the symmetric minimum occurs at

$a=b=-\frac1{\sqrt2}.$

Hence

$\min \left(2^{\sin x}+2^{\cos x}\right)=2\cdot 2^{-1/\sqrt2}=2^{1-1/\sqrt2}.$

5. Therefore the correct option is

$\boxed{2^{1-\frac1{\sqrt2}}}.$