Solutions of tan x equals tan a/Examples/2 sec squared x equals 5 tan x
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Examples of Use of Solutions of $\tan x = \tan a$
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.
Proof
| \(\ds 2 \sec^2 x\) | \(=\) | \(\ds 5 \tan x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 \paren {1 + \tan^2 x}\) | \(=\) | \(\ds 5 \tan x\) | Difference of Squares of Secant and Tangent | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 \tan^2 x - 5 \tan x + 2\) | \(=\) | \(\ds 0\) | rearranging | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {2 \tan x - 1} \paren {\tan x - 2}\) | \(=\) | \(\ds 0\) | rearranging | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \tan x\) | \(\in\) | \(\ds \set {\dfrac 1 2, 2}\) | equating factors | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds n \pi + \arctan \dfrac 1 2\) | Solutions of $\tan x = \tan a$ | ||||||||||
| \(\, \ds \lor \, \) | \(\ds x\) | \(=\) | \(\ds n \pi + \arctan 2\) |
Hence the result.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Solution of equations: Example $2$.