The maximum value of 2 sinx+ cos2x attains at x =

$$\begin{gathered} \text{⨍ (x)=2sinx+cos2x} \\ {\text{⨍ }}^1\text{(x)=2cosx-2sin2x} \end{gathered}$$

$\text{⨍ (x)}$ is min or max

$$\begin{gathered} {\text{⨍ }}^1\text{(x)=0} \\ 2\mathrm{cosx}-4\mathrm{sinx} \mathrm{cosx}=0 \\ 2\mathrm{cosx}(1-2\mathrm{sinx})=0 \\ \mathrm{cosx}=0 \mathrm{sinx}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{x}=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \mathrm{x}=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \\ \text{⨍ (}\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{)=2-1=1} \\ \text{⨍ }\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)\text{=2.}\frac{1}{2}\text{+}\frac{1}{2}\text{=}\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$

at $\mathrm{x}=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2 \text{⨍ (x)}$}\right.$ is min

at $\mathrm{x}=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \text{⨍ (x)}$ is max