Determine the angles that satisfy the $ an$ function

To determine the angles $\theta$ where $\tan(\theta) = \sqrt{3}$, we first recognize that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. From the unit circle, we know:

$ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} $

Thus, $\theta = \frac{\pi}{3}$ is one solution. Since tangent has a period of $\pi$, the general solution can be written as:

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

where $n$ is any integer.

Another solution within one period (from

txt1

txt1

txt1

$ to $2\pi$) would be:

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

Therefore, the angles that satisfy $\tan(\theta) = \sqrt{3}$ within one period are $\theta = \frac{\pi}{3}$ and $\theta = \frac{4\pi}{3}$.

To find the angles $ heta$ where $ an( heta) = sqrt{3}$, we need to recall the values from the unit circle. The tangent function is the ratio of the sine and cosine functions:

$ an( heta) = frac{sin( heta)}{cos( heta)} $

From the unit circle, we see:

$ anleft(frac{pi}{3}
ight) = sqrt{3} $

Thus, one solution is:

$ heta = frac{pi}{3} $

Since tangent repeats every $pi$ radians, the general solution is:

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

where $k$ is any integer.

Another solution within one period (

txt2

txt2

txt2

$ to $2pi$) is:

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

In summary, the angles that satisfy $ an( heta) = sqrt{3}$ within one period are $ heta = frac{pi}{3}$ and $ heta = frac{4pi}{3}$.

To solve $ an( heta) = sqrt{3}$, we refer to the unit circle. We know:

$ anleft(frac{pi}{3}
ight) = sqrt{3} $

Thus, $ heta = frac{pi}{3}$ and $ heta = frac{4pi}{3}$ within one period satisfy this equation.

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