I think you just need to know your unit circle and values for the sin and cos. The only other way I can think of is trying to solve for pi(t) via inverse trig functions but I'm not getting any simplifications from Mathematica using that tactic.
If you want you could solve it algebraically by squaring both sides of cos x = -sin x and then expressing cos^2 x in terms of sin^2 x, then rearranging terms and using an inverse function, but that's more complicated than you need and it might make you miss one of the solutions if you're not careful. You can solve this one just by looking at the unit circle.
Or you may try to do the following:
If [tex]\cos (\pi (t)) = 0[/tex] then [tex]\sin (\pi (t)) = \pm 1[/tex]
So [tex]\cos (\pi (t)) + \sin (\pi (t)) = 0 \pm 1 = \pm 1 \neq 0[/tex]
That means if [tex]\cos (\pi (t)) = 0[/tex] then the LHS is not 0, and hence it does not satify the equation.
So it's true that [tex]\cos (\pi (t)) \neq 0[/tex].
Divide both sides of the equation by [tex]\cos (\pi (t))[/tex] to obtain:
[tex]1 + \tan (\pi (t)) = 0[/tex]
Now, can you go from here? :)