If `2sin^(2)theta+sqrt(3)costheta+1=0`, then the value of `theta` is

Options

A. `(pi)/(6)`
B. `(2pi)/(3)`
C. `(5pi)/(6)`
D. `pi`

Explanation

To solve the equation \(2\sin^2\theta + \sqrt{3}\cos\theta + 1 = 0\), we will follow these steps:

### Step 1: Convert \(\sin^2\theta\) to \(\cos\theta\)
We know that:
\[ \sin^2\theta = 1 - \cos^2\theta \]
Substituting this into the equation gives:
\[ 2(1 - \cos^2\theta) + \sqrt{3}\cos\theta + 1 = 0 \]
This simplifies to:
\[ 2 - 2\cos^2\theta + \sqrt{3}\cos\theta + 1 = 0 \]

### Step 2: Rearranging the equation
Combine like terms:
\[ 3 - 2\cos^2\theta + \sqrt{3}\cos\theta = 0 \]
Rearranging gives:
\[ -2\cos^2\theta + \sqrt{3}\cos\theta + 3 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ 2\cos^2\theta - \sqrt{3}\cos\theta - 3 = 0 \]

### Step 3: Use the quadratic formula
This is a quadratic equation in terms of \(\cos\theta\). We can use the quadratic formula:
\[ \cos\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 2\), \(b = -\sqrt{3}\), and \(c = -3\). Plugging in these values:
\[ \cos\theta = \frac{\sqrt{3} \pm \sqrt{(-\sqrt{3})^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ \cos\theta = \frac{\sqrt{3} \pm \sqrt{3 + 24}}{4} = \frac{\sqrt{3} \pm \sqrt{27}}{4} = \frac{\sqrt{3} \pm 3\sqrt{3}}{4} \]

### Step 4: Simplifying the solutions
This gives us two potential solutions:
1. \(\cos\theta = \frac{\sqrt{3} + 3\sqrt{3}}{4} = \frac{4\sqrt{3}}{4} = \sqrt{3}\)
2. \(\cos\theta = \frac{\sqrt{3} - 3\sqrt{3}}{4} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}\)

### Step 5: Analyzing the solutions
The first solution, \(\cos\theta = \sqrt{3}\), is not valid since the range of \(\cos\theta\) is \([-1, 1]\).

The second solution, \(\cos\theta = -\frac{\sqrt{3}}{2}\), is valid. The angles that satisfy this equation are:
\[ \theta = \frac{5\pi}{6} \quad \text{and} \quad \theta = \frac{7\pi}{6} \]

### Final Answer
Thus, the value of \(\theta\) is:
\[ \theta = \frac{5\pi}{6} \quad \text{(in the range of } [0, 2\pi]\text{)} \]