Solve for integers

Dec 2015
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116
Earth
\(\displaystyle \sin(\pi k) +\cos(\pi k) =1\)
k-integer.
 

romsek

Math Team
Sep 2015
2,689
1,489
USA
$\sin(\pi k)=0,~k\in \mathbb{Z}$

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

$\text{so $\sin(\pi k)+\cos(\pi k)=1,~\forall k \in \mathbb{Z}\wedge k$ is even}$
 
Jun 2019
455
237
USA
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
Dec 2006
21,205
2,337
It also gives the solutions for the corresponding equation with "-1" instead of "1". That could be considered a bonus rather than an error.
 
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