# Solve for integers

#### idontknow

$$\displaystyle \sin(\pi k) +\cos(\pi k) =1$$
k-integer.

#### romsek

Math Team
$\sin(\pi k)=0,~k\in \mathbb{Z}$

$\cos(\pi k)=1,~\forall k \text{ even}$

$\text{so$\sin(\pi k)+\cos(\pi k)=1,~\forall k \in \mathbb{Z}\wedge k$is even}$

#### tahirimanov19

Take the square...

#### DarnItJimImAnEngineer

What good does taking the square do? It gives you $2\cos(\pi k)\sin(\pi k)=0$, which erroneously suggests a solution $k \in \mathbb{Z}$ instead of $k \in 2\mathbb{Z}$.

#### skipjack

Forum Staff
It also gives the solutions for the corresponding equation with "-1" instead of "1". That could be considered a bonus rather than an error.

topsquark

#### DarnItJimImAnEngineer

"It's a feature!"

topsquark
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