Question

Minimum value of 2^sinx+2^cosx

Answer 1

Step-by-step explanation:

\begin{lgathered}2 \sin(x) + 3 \cos(x ) = 2 \\ 2 \sin(x) - 3 \cos(x) = 1 \div 2 \\ 4 \sin(x ) = 5 \div 2 \\ \sin(x) = 5 \div 8 \\ 6 \cos(x ) = 3 \div 2 \\ \cos(x) = 1 \div 4\end{lgathered}$$

Step-by-step explanation:

so keep sinx and cosx values in required equation

$$3(5 \div 8) + 2(1 \div 4) = 19 \div 8$$

Answer 2

Step-by-step explanation:

the minimum value of sinx is 0

angle is 0°, cos0°is 1

so minimum value of 2^sinx+2^cosx is 2^0 +2^1=1+2

=3