Let `0lt theta_(1) lt theta_(2) lt theta_(3) lt……….` denotes the positive solutions of the equation `3+3 cos theta =2sin^(2)theta.` If `theta_(3)+theta_(7)=api`, where a is an integer, then the value of a is equal to

Options

A. 6
B. 7
C. 8
D. 4

Answer

To solve the equation \(3 + 3 \cos \theta = 2 \sin^2 \theta\) and find the integer \(a\) such that \(\theta_3 + \theta_7 = a\pi\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 3 + 3 \cos \theta = 2 \sin^2 \theta \]
Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\), we can rewrite the equation:
\[ 3 + 3 \cos \theta = 2(1 - \cos^2 \theta) \]
This simplifies to:
\[ 3 + 3 \cos \theta = 2 - 2 \cos^2 \theta \]

### Step 2: Rearranging the equation
Rearranging gives:
\[ 2 \cos^2 \theta + 3 \cos \theta + 1 = 0 \]

### Step 3: Solving the quadratic equation
Now we can treat this as a quadratic equation in terms of \(\cos \theta\):
\[ 2x^2 + 3x + 1 = 0 \]
where \(x = \cos \theta\). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 2\), \(b = 3\), and \(c = 1\):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm 1}{4} \]
This gives us two solutions:
\[ x_1 = \frac{-2}{4} = -\frac{1}{2}, \quad x_2 = \frac{-4}{4} = -1 \]

### Step 4: Finding angles
Now we find \(\theta\) for these values:
1. For \(\cos \theta = -\frac{1}{2}\):
\[ \theta = \frac{2\pi}{3} + 2n\pi \quad \text{and} \quad \theta = \frac{4\pi}{3} + 2n\pi \]
2. For \(\cos \theta = -1\):
\[ \theta = \pi + 2n\pi \]

### Step 5: Listing positive solutions
The positive solutions in increasing order are:
- For \(n = 0\):
- \(\frac{2\pi}{3}\)
- \(\frac{4\pi}{3}\)
- \(\pi\) (from \(\cos \theta = -1\))

Continuing with \(n = 1\):
- \(\frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}\)
- \(\frac{4\pi}{3} + 2\pi = \frac{10\pi}{3}\)
- \(\pi + 2\pi = 3\pi\)

Continuing with \(n = 2\):
- \(\frac{2\pi}{3} + 4\pi = \frac{14\pi}{3}\)
- \(\frac{4\pi}{3} + 4\pi = \frac{16\pi}{3}\)
- \(\pi + 4\pi = 5\pi\)

### Step 6: Identifying \(\theta_3\) and \(\theta_7\)
The positive solutions in order are:
1. \(\frac{2\pi}{3}\) (1st)
2. \(\pi\) (2nd)
3. \(\frac{4\pi}{3}\) (3rd)
4. \(3\pi\) (4th)
5. \(\frac{8\pi}{3}\) (5th)
6. \(\frac{10\pi}{3}\) (6th)
7. \(\frac{14\pi}{3}\) (7th)

Thus, we have:
\[ \theta_3 = \frac{4\pi}{3}, \quad \theta_7 = \frac{14\pi}{3} \]

### Step 7: Calculating \(\theta_3 + \theta_7\)
Now we compute:
\[ \theta_3 + \theta_7 = \frac{4\pi}{3} + \frac{14\pi}{3} = \frac{18\pi}{3} = 6\pi \]

### Step 8: Finding \(a\)
Since \(\theta_3 + \theta_7 = a\pi\), we have:
\[ 6\pi = a\pi \implies a = 6 \]

Thus, the value of \(a\) is:
\[ \boxed{6} \]