Is it valid to write the solution to an equation like that?

Question

I solved a trigonometric equation and got the following solutions: $$x=\frac{\pi}{6}+2\pi n$$ $$x=\frac{5\pi}{6}+2\pi n$$ $$x=-\frac{\pi}{2}+2\pi n$$

For all integer $n$.

Is it valid to write the solution like that?

$$x \in \{ n\in \mathbb{Z} \ | \ (x=\frac{\pi}{6}+2\pi n \ \cup x=\frac{5\pi}{6}+2\pi n \ \cup x=-\frac{\pi}{2}+2\pi n)\}$$

Answer 1

No. Those $\cup$ signs make no sense and you are asserting that your set of solutions is a subset of $\mathbb Z$.

You can write your set as$$\left\{\frac\pi6+2\pi n\,\middle|\,n\in\mathbb Z\right\}\cup\left\{\frac{5\pi}6+2\pi n\,\middle|\,n\in\mathbb Z\right\}\cup\left\{-\frac\pi2+2\pi n\,\middle|\,n\in\mathbb Z\right\}.$$

Answer 2

You should use "logical or" connectors $\lor$. But what about

$$\frac{4n+1}6\pi\ ?$$

Answer 3

You can't: in the first place, the solutions are not integers. Personally, I would use congruences: $$x\equiv \frac\pi 6,\:\frac{5\pi}6,\:-\frac\pi 2\pmod{2\pi}$$ or, with sets, $$x\in\Bigl(\frac\pi 6+2\pi \mathbf Z\Bigr)\cup \Bigl(\frac{5\pi}6+2\pi \mathbf Z\Bigr)\cup\Bigl(-\frac\pi 2+2\pi \mathbf Z\Bigr).$$