Number of roots of `cos^2x+(sqrt(3)+1)/2sinx-(sqrt(3))/4-1=0` which lie in the interval `[-pi,pi]` is 2 (b) 4 (c) 6 (d) 8

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To find the number of roots of the equation

\[ \cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0 \]

that lie in the interval \([-π, π]\), we will follow these steps:

### Step 1: Rewrite the equation
Start by rewriting \(\cos^2 x\) in terms of \(\sin x\):
\[ \cos^2 x = 1 - \sin^2 x \]
Substituting this into the equation gives:
\[ 1 - \sin^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0 \]

### Step 2: Simplify the equation
The \(1\) and \(-1\) cancel out:
\[ -\sin^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} = 0 \]
Multiplying through by \(-1\) to make the leading coefficient positive:
\[ \sin^2 x - \frac{\sqrt{3} + 1}{2} \sin x + \frac{\sqrt{3}}{4} = 0 \]

### Step 3: Use the quadratic formula
This is a quadratic equation in \(\sin x\). We can use the quadratic formula:
\[ \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -\frac{\sqrt{3} + 1}{2}\), and \(c = \frac{\sqrt{3}}{4}\).

Calculating the discriminant:
\[ b^2 - 4ac = \left(-\frac{\sqrt{3} + 1}{2}\right)^2 - 4 \cdot 1 \cdot \frac{\sqrt{3}}{4} \]
\[ = \frac{(\sqrt{3} + 1)^2}{4} - \sqrt{3} \]
\[ = \frac{3 + 2\sqrt{3} + 1}{4} - \sqrt{3} \]
\[ = \frac{4 + 2\sqrt{3}}{4} - \frac{4\sqrt{3}}{4} \]
\[ = \frac{4 - 2\sqrt{3}}{4} \]

### Step 4: Solve for \(\sin x\)
Now substituting back into the quadratic formula:
\[ \sin x = \frac{\frac{\sqrt{3} + 1}{2} \pm \sqrt{\frac{4 - 2\sqrt{3}}{4}}}{2} \]
\[ = \frac{\sqrt{3} + 1 \pm \sqrt{4 - 2\sqrt{3}}}{4} \]

### Step 5: Find the angles
We need to find the angles \(x\) corresponding to the values of \(\sin x\) found in the previous step. The sine function is periodic and has specific values in the intervals:
- For \(\sin x = \frac{1}{2}\), the angles are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\).
- For \(\sin x = \frac{\sqrt{3}}{2}\), the angles are \(x = \frac{\pi}{3}\) and \(x = \frac{2\pi}{3}\).

### Step 6: Count the roots
In the interval \([-π, π]\), we also consider negative angles:
- For \(\sin x = \frac{1}{2}\): \(x = -\frac{5\pi}{6}\) and \(x = -\frac{\pi}{6}\).
- For \(\sin x = \frac{\sqrt{3}}{2}\): \(x = -\frac{2\pi}{3}\) and \(x = -\frac{\pi}{3}\).

Thus, we have a total of 4 roots:
1. \(\frac{\pi}{6}\)
2. \(\frac{5\pi}{6}\)
3. \(-\frac{\pi}{6}\)
4. \(-\frac{5\pi}{6}\)

### Conclusion
The total number of roots of the equation in the interval \([-π, π]\) is **4**.