General Solution for tan x = 1/√3

The question asks us to find the general solution for the trigonometric equation given as:

\[ \tan x = \frac{1}{\sqrt{3}} \]

Understanding the Tangent Value

To find the general solution, we first need to determine the principal value of $x$ for which the tangent function equals $\frac{1}{\sqrt{3}}$. We recall the standard trigonometric values for common angles:

• We know that $\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$. This means that when the angle is $\frac{\pi}{6}$ radians (or 30 degrees), its tangent is $\frac{1}{\sqrt{3}}$.

Therefore, we can rewrite our original equation in the form $\tan x = \tan \alpha$:

\[ \tan x = \tan \left(\frac{\pi}{6}\right) \]

General Solution Formula for Tangent Equations

For any trigonometric equation of the form $\tan x = \tan \alpha$, where $\alpha$ is a known angle, the general solution is given by the formula:

\[ x = n\pi + \alpha \]

Here, $n$ represents any integer (i.e., $n \in \mathbb{Z}$). This formula accounts for all possible angles that have the same tangent value because the tangent function has a period of $\pi$. This means its values repeat every $\pi$ radians.

Applying the General Solution Formula

In our specific equation, we have identified that $\alpha = \frac{\pi}{6}$. Now, we substitute this value into the general solution formula for tangent:

\[ x = n\pi + \frac{\pi}{6} \]

This expression provides all possible values of $x$ that satisfy the original trigonometric equation $\tan x = \frac{1}{\sqrt{3}}$. For example, if $n=0$, $x = \frac{\pi}{6}$; if $n=1$, $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$; if $n=-1$, $x = -\pi + \frac{\pi}{6} = -\frac{5\pi}{6}$, and so on. All these values of $x$ will have a tangent equal to $\frac{1}{\sqrt{3}}$.

Final General Solution

The general solution for the equation $\tan x = \frac{1}{\sqrt{3}}$ is:

\[ x = n\pi + \frac{\pi}{6} \]

where $n$ is an integer.