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To solve for $\theta$ in $\cot(\theta) = \frac{1}{\sqrt{3}}$, we start by recalling the definition of cotangent:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Given $\cot(\theta) = \frac{1}{\sqrt{3}}$, we have:

$\frac{1}{\tan(\theta)} = \frac{1}{\sqrt{3}}$

By taking reciprocals, we find:

$\tan(\theta) = \sqrt{3}$

The angles that satisfy $\tan(\theta) = \sqrt{3}$ on the unit circle are:

$\theta = \frac{\pi}{3} + k\pi$

where $k$ is any integer. Considering the interval $[0, 2\pi)$, we get:

$\theta = \frac{\pi}{3}, \frac{4\pi}{3}$

So, the angles are:

$\boxed{\frac{\pi}{3}, \frac{4\pi}{3}}$

To solve for $ heta$ where $cot( heta) = frac{1}{sqrt{3}}$, we use the relationship between cotangent and tangent:

$cot( heta) = frac{1}{ an( heta)}$

Given $cot( heta) = frac{1}{sqrt{3}}$, we find:

$ an( heta) = sqrt{3}$

The values of $ heta$ that satisfy $ an( heta) = sqrt{3}$ are:

$ heta = frac{pi}{3} + kpi$

where $k$ is an integer. For the interval $[0, 2pi)$, we have two solutions:

$ heta = frac{pi}{3}, frac{4pi}{3}$

Thus, the angles are:

$oxed{frac{pi}{3}, frac{4pi}{3}}$

Find $ heta$ such that $cot( heta) = frac{1}{sqrt{3}}$ in $[0, 2pi)$. Since $cot( heta) = frac{1}{ an( heta)}$, this implies:

$ an( heta) = sqrt{3}$

Thus:

$ heta = frac{pi}{3}, frac{4pi}{3}$

Therefore:

$oxed{frac{pi}{3}, frac{4pi}{3}}$

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