## Precalculus (6th Edition)

$\{180^{o}+k\cdot 360^{o},$ k any integer$\}$
$\cos\theta=-1$ 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$\}$