Solutions of tan x equals tan a/Proof

Theorem

Let $\alpha \in \R$ be fixed.

Let:

$(1): \quad \tan x = \tan \alpha$

The solution set of $(1)$ is:

$\set {x \in \R: \forall n \in \Z: x = n \pi + \alpha}$

Proof

From Tangent Function is Periodic on Reals:

$\map \tan {\pi + x} = \tan x$

Hence:

\(\ds x\)

\(=\)

\(\ds n \pi + a\)

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: General solution of $\tan x = \tan \alpha$