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