If the sum of the root of the equation ` cos 4x + 6+7 cos 2x ` in the interval [0,314] is `kpi, k in R ` Find (k-1248)

Answer

To solve the equation \( \cos 4x + 6 + 7 \cos 2x = 0 \) in the interval \([0, 314]\) and find \(k\) such that the sum of the roots is \(k\pi\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos 4x + 6 + 7 \cos 2x = 0 \]
Using the double angle identity for cosine, we know that:
\[ \cos 4x = 2\cos^2 2x - 1 \]
Substituting this into the equation gives:
\[ 2\cos^2 2x - 1 + 6 + 7 \cos 2x = 0 \]
This simplifies to:
\[ 2\cos^2 2x + 7\cos 2x + 5 = 0 \]

### Step 2: Let \(t = \cos 2x\)
Let \(t = \cos 2x\). The equation now becomes:
\[ 2t^2 + 7t + 5 = 0 \]

### Step 3: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 2\), \(b = 7\), and \(c = 5\):
\[ t = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 7^2 - 4 \cdot 2 \cdot 5 = 49 - 40 = 9 \]
So, we have:
\[ t = \frac{-7 \pm 3}{4} \]
Calculating the two possible values for \(t\):
1. \(t = \frac{-7 + 3}{4} = \frac{-4}{4} = -1\)
2. \(t = \frac{-7 - 3}{4} = \frac{-10}{4} = -\frac{5}{2}\)

### Step 4: Determine valid solutions for \(t\)
The value \(t = \cos 2x\) must be in the range \([-1, 1]\). Therefore, \(t = -\frac{5}{2}\) is not valid. We only consider:
\[ t = -1 \]

### Step 5: Solve for \(x\)
If \(t = -1\), then:
\[ \cos 2x = -1 \]
This occurs at:
\[ 2x = (2n + 1)\pi \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ x = \frac{(2n + 1)\pi}{2} \]

### Step 6: Find the values of \(n\) in the interval
We need to find \(n\) such that:
\[ 0 \leq \frac{(2n + 1)\pi}{2} \leq 314 \]
This simplifies to:
\[ 0 \leq 2n + 1 \leq \frac{628}{\pi} \]
Calculating \(\frac{628}{\pi}\):
\[ \frac{628}{\pi} \approx 200 \]
Thus:
\[ 0 \leq 2n + 1 \leq 200 \implies -1 \leq 2n \leq 199 \implies 0 \leq n \leq 99 \]
So \(n\) can take values from \(0\) to \(99\), giving us \(100\) possible values.

### Step 7: Calculate the sum of the roots
The roots are:
\[ x = \frac{1\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots, \frac{199\pi}{2} \]
This is an arithmetic series with:
- First term \(a = \frac{\pi}{2}\)
- Last term \(l = \frac{199\pi}{2}\)
- Number of terms \(n = 100\)

The sum \(S_n\) of an arithmetic series is given by:
\[ S_n = \frac{n}{2} (a + l) \]
Substituting the values:
\[ S_{100} = \frac{100}{2} \left(\frac{\pi}{2} + \frac{199\pi}{2}\right) = 50 \left(\frac{200\pi}{2}\right) = 50 \cdot 100\pi = 5000\pi \]

### Step 8: Find \(k\)
We have:
\[ k\pi = 5000\pi \implies k = 5000 \]

### Step 9: Calculate \(k - 1248\)
Finally, we need to find:
\[ k - 1248 = 5000 - 1248 = 3752 \]

Thus, the final answer is:
\[ \boxed{3752} \]