If the points `(-2, 0), (-1,(1)/(sqrt(3)))` and `(cos theta, sin theta)` are collinear, then the number of value of `theta in [0, 2pi]` is

Answer

To determine the number of values of \( \theta \) in the interval \([0, 2\pi]\) such that the points \((-2, 0)\), \((-1, \frac{1}{\sqrt{3}})\), and \((\cos \theta, \sin \theta)\) are collinear, we can follow these steps:

### Step 1: Set up the condition for collinearity
The points are collinear if the area of the triangle formed by them is zero. The area \( A \) of a triangle given by vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) can be calculated using the determinant formula:

\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

For our points, we have:
- \((x_1, y_1) = (-2, 0)\)
- \((x_2, y_2) = (-1, \frac{1}{\sqrt{3}})\)
- \((x_3, y_3) = (\cos \theta, \sin \theta)\)

Setting the area to zero gives us:

\[ -2\left(\frac{1}{\sqrt{3}} - \sin \theta\right) + (-1)(\sin \theta - 0) + \cos \theta\left(0 - \frac{1}{\sqrt{3}}\right) = 0 \]

### Step 2: Simplify the equation
Expanding and simplifying the equation:

\[ -2\left(\frac{1}{\sqrt{3}} - \sin \theta\right) - \sin \theta - \frac{\cos \theta}{\sqrt{3}} = 0 \]

This simplifies to:

\[ -2\frac{1}{\sqrt{3}} + 2\sin \theta - \sin \theta - \frac{\cos \theta}{\sqrt{3}} = 0 \]

Combining like terms:

\[ \sin \theta - 2\frac{1}{\sqrt{3}} - \frac{\cos \theta}{\sqrt{3}} = 0 \]

### Step 3: Rearranging the equation
Rearranging gives:

\[ \sin \theta = 2\frac{1}{\sqrt{3}} + \frac{\cos \theta}{\sqrt{3}} \]

### Step 4: Multiply through by \(\sqrt{3}\)
To eliminate the fraction, multiply the entire equation by \(\sqrt{3}\):

\[ \sqrt{3} \sin \theta = 2 + \cos \theta \]

### Step 5: Use trigonometric identities
Using the sine subtraction formula, we can express this equation as:

\[ \sqrt{3} \sin \theta - \cos \theta = 2 \]

This can be rewritten using the sine subtraction identity:

\[ \sin\left(\theta - \frac{\pi}{6}\right) = 1 \]

### Step 6: Solve for \(\theta\)
The sine function equals 1 at:

\[ \theta - \frac{\pi}{6} = \frac{\pi}{2} + 2k\pi \quad (k \in \mathbb{Z}) \]

Thus:

\[ \theta = \frac{\pi}{2} + \frac{\pi}{6} + 2k\pi \]

Calculating this gives:

\[ \theta = \frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} + 2k\pi \]

### Step 7: Find values in the interval \([0, 2\pi]\)
For \( k = 0 \):

\[ \theta = \frac{2\pi}{3} \]

For \( k = 1 \):

\[ \theta = \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3} \quad \text{(not in the interval)} \]

Thus, the only valid solution in the interval \([0, 2\pi]\) is:

\[ \theta = \frac{2\pi}{3} \]

### Conclusion
The number of values of \( \theta \) in the interval \([0, 2\pi]\) for which the points are collinear is **1**.

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