The polar coordinates equivalent to `(-3,sqrt3)` are

Options

A. `(2sqrt3,(pi)/(6))`
B. `(-2sqrt3,(5pi)/(6))`
C. `(2sqrt3,(7pi)/(6))`
D. `(2sqrt3,(5pi)/(6))`

Explanation

To find the polar coordinates equivalent to the Cartesian coordinates \((-3, \sqrt{3})\), we will follow these steps:

### Step 1: Calculate the value of \( r \)
The polar coordinate \( r \) is calculated using the formula:
\[ r = \sqrt{x^2 + y^2} \]
Here, \( x = -3 \) and \( y = \sqrt{3} \).

Substituting the values:
\[ r = \sqrt{(-3)^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3} \]

### Step 2: Calculate the angle \( \theta \)
The angle \( \theta \) can be calculated using the formula:
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
However, since \( x \) is negative and \( y \) is positive, the point \((-3, \sqrt{3})\) lies in the second quadrant. We will first find the reference angle and then adjust it for the second quadrant.

Calculating the reference angle:
\[ \theta_{\text{ref}} = \tan^{-1}\left(\frac{\sqrt{3}}{-3}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \]
This gives us:
\[ \theta_{\text{ref}} = -\frac{\pi}{6} \]

To find the angle in the second quadrant, we add \(\pi\):
\[ \theta = \pi - \left(-\frac{\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{6\pi}{6} + \frac{\pi}{6} = \frac{7\pi}{6} \]

### Step 3: Write the polar coordinates
Now that we have both \( r \) and \( \theta \), we can express the polar coordinates:
\[ (r, \theta) = (2\sqrt{3}, \frac{7\pi}{6}) \]

### Final Answer
The polar coordinates equivalent to \((-3, \sqrt{3})\) are:
\[ (2\sqrt{3}, \frac{7\pi}{6}) \]