#### Answer

$\{180^{o}+k\cdot 360^{o}, $ k any integer$\}$

#### Work Step by Step

$\cos\theta=-1$
In the interval $[0,360^{o})$, a solution is
$\theta=180^{o}$
Since the period for cosine is $360^{o}$, then all angles
$180^{o}+k\cdot 360^{o}, $ k any integer,
are also solutions.
Solution set:
$\{180^{o}+k\cdot 360^{o}, $ k any integer$\}$