#### Answer

$\{135^{o}+k\cdot 180^{o}, $ k any integer$\}$

#### Work Step by Step

$\tan\theta=-1$
In the interval $[0,360^{o})$, a solution is
$\theta=135^{o}$
Since the period for tan is $180^{o}$, then all angles
$135^{o}+k\cdot 180^{o}, $ k any integer,
are also solutions.
Solution set:
$\{135^{o}+k\cdot 180^{o}, $ k any integer$\}$