If i have this exercise:
$\tan(\alpha) = 0$, the solution is when the $\sin(a) = 0$, that is in $0°$ and $180°$, because $\tan = \sin/\cos$
So the solution must be: $\alpha_1 = 0 + \pi k$, $\alpha_2 = \pi + \pi k$
But according to symbolab, the unique solution is the $\alpha_1$, so why $\alpha_2$ is incorrect?
Note that $$\pi k $$ also covers $$\pi + \pi k $$ because $$\pi + \pi k = \pi ( k+1) $$ and as $ k$ runs over integers, it covers $k+1$ as well.