To solve the equation where the tangent of θ equals the square root of 3, we first identify the angles on the unit circle that satisfy this condition. The tangent function, defined as the ratio of the sine to the cosine (i.e., $\backslash tan(\backslash theta)\; =\; \backslash frac\{\backslash sin(\backslash theta)\}\{\backslash cos(\backslash theta)\}$), is positive in the first and third quadrants.

In the first quadrant, the angle $\backslash frac\{\backslash pi\}\{3\}$ has a tangent value of $\backslash sqrt\{3\}$. In the third quadrant, the angle $\backslash frac\{4\backslash pi\}\{3\}$ also yields a tangent of $\backslash sqrt\{3\}$. Thus, the two primary solutions we find are $\backslash frac\{\backslash pi\}\{3\}$ and $\backslash frac\{4\backslash pi\}\{3\}$.

To express all possible solutions, we add $2\backslash pi\; n$ to each of these angles, where $n$ is any integer, resulting in the general solutions:

• $\backslash theta\; =\; \backslash frac\{\backslash pi\}\{3\}\; +\; 2\backslash pi\; n$
• $\backslash theta\; =\; \backslash frac\{4\backslash pi\}\{3\}\; +\; 2\backslash pi\; n$

However, since $\backslash frac\{\backslash pi\}\{3\}$ and $\backslash frac\{4\backslash pi\}\{3\}$ are exactly $\backslash pi$ radians apart, we can simplify our solution further. By starting with $\backslash frac\{\backslash pi\}\{3\}$ and adding $\backslash pi\; n$ (which accounts for all angles that are $\backslash pi$ radians apart), we can express the complete set of solutions as:

$\backslash theta\; =\; \backslash frac\{\backslash pi\}\{3\}\; +\; \backslash pi\; n$

This formulation captures all solutions to the equation $\backslash tan(\backslash theta)\; =\; \backslash sqrt\{3\}$ in its most simplified form. It is important to note that this simplification is specific to the tangent function, as it exhibits periodicity with a period of $\backslash pi$.