14. Given that $y = f ( x )$ is a function with period $2 \pi$, and when $x \in [ 0,2 \pi )$, $f ( x ) = \sin \frac { x } { 2 }$, the solution set of $f ( x ) = \frac { 1 } { 2 }$ is A. $\left\{ x \left\lvert \, x = 2 k \pi + \frac { \pi } { 3 } \right. , k \in \mathbb{Z} \right\}$. B. $\left\{ x \left\lvert \, x = 2 k \pi + \frac { 5 \pi } { 3 } \right. , k \in \mathbb{Z} \right\}$. C. $\left\{ x \left\lvert \, x = 2 k \pi \pm \frac { \pi } { 3 } \right. , k \in \mathbb{Z} \right\}$. D. $\left\{ x \left\lvert \, x = 2 k \pi + \frac { \pi } { 3 } + ( - 1 ) ^ { k } \frac{\pi}{3} \right. , k \in \mathbb{Z} \right\}$.
14. Given that $y = f ( x )$ is a function with period $2 \pi$, and when $x \in [ 0,2 \pi )$, $f ( x ) = \sin \frac { x } { 2 }$, the solution set of $f ( x ) = \frac { 1 } { 2 }$ is\\
A. $\left\{ x \left\lvert \, x = 2 k \pi + \frac { \pi } { 3 } \right. , k \in \mathbb{Z} \right\}$.\\
B. $\left\{ x \left\lvert \, x = 2 k \pi + \frac { 5 \pi } { 3 } \right. , k \in \mathbb{Z} \right\}$.\\
C. $\left\{ x \left\lvert \, x = 2 k \pi \pm \frac { \pi } { 3 } \right. , k \in \mathbb{Z} \right\}$.\\
D. $\left\{ x \left\lvert \, x = 2 k \pi + \frac { \pi } { 3 } + ( - 1 ) ^ { k } \frac{\pi}{3} \right. , k \in \mathbb{Z} \right\}$.