The minimum value of $$3^{sinx}+3^{cosx}$$ is

Using AM $$\geq$$ GM, We can say that 

$$\dfrac{3^{sinx}+3^{cosx}}{2}$$ $$\geq$$ $$\sqrt{3^{sinx}*3^{cosx}}$$

$$\Rightarrow$$ $$3^{sinx}+3^{cosx}$$ $$\geq$$ $$2*3^\frac{sinx+cosx}{2}$$  ... (1)

We know that -$$\sqrt{A^2+B^2}$$ $$\leq$$ Asinx+Bcosx $$\leq$$ $$\sqrt{A^2+B^2}$$

Therefore,   -$$\sqrt{1^2+1^2}$$ $$\leq$$ sinx+cosx $$\leq$$ $$\sqrt{1^2+1^2}$$

$$\Rightarrow$$ -$$\sqrt{2}$$ $$\leq$$ sinx+cosx $$\leq$$ $$\sqrt{2}$$

Hence, we can say that the minimum value of sinx+cosx = -$$\sqrt{2}$$  ... (2)

From equation (1) and (2) we can say that,

$$\Rightarrow$$ $$3^{sinx}+3^{cosx}$$ $$\geq$$ $$2*3^\frac{-\sqrt{2}}{2}$$

$$\Rightarrow$$ $$3^{sinx}+3^{cosx}$$ $$\geq$$ $$2*3^\frac{-1}{\sqrt{2}}$$

Therefore, option B is the correct answer.