If $\cos \theta =-\frac{1}{2}$ and $0<\theta <{360}^{\circ }$ ,then the solutions are

The correct option is B. $\theta ={120}^{\circ },{240}^{\circ }$

Given: $\cos \theta =-\frac{1}{2}$

$\cos \theta =-\cos {60}^{\circ }$

Since, $\cos \theta <0$ in second and third quadrants, and

$\cos ({180}^{\circ }-\theta )=\cos ({180}^{\circ }+\theta )=-\cos \theta$

$\Rightarrow \theta ={180}^{\circ }-{60}^{\circ }$ and ${180}^{\circ }+{60}^{\circ }$

$\Rightarrow \theta ={120}^{\circ }$ and ${240}^{\circ }$.