cho cosx =-1/2 (0°<x<180°)

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
cosx\ =\ -\frac{1}{2}\\
\Longrightarrow \ x\ =\ \frac{2\pi }{3} \ +\ k2\pi 
\end{array}$

Mà $\displaystyle 0^{0} < x< 180^{0}$

nên $\displaystyle x\ =\ \frac{2\pi }{3}$