Solutions of tan x equals tan a

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:

\(\ds x\)

\(=\)

\(\ds n \pi + a\)

$\blacksquare$

Examples

Solutions of $2 \sec^2 x = 5 \tan x$

The equation

$2 \sec^2 x = 5 \tan x$

has the general solution:

$\set {n \pi + \arctan \dfrac 1 2: n \in \Z} \cup \set {n \pi + \arctan 2: n \in \Z}$

where $\arctan$ denotes the (real) arctangent function.

Solutions of $\tan^2 x - 3 \tan x + 2 = 0$

Solutions of tan x equals tan a/Examples/tan squared x minus 3 tan x plus 2 equals 0

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(36)$