If `a^(2)-2a cos x +1=674` and `tan (x/2)=7` then the integral value of a is

Answer

To solve the problem step by step, we start with the given equation and the information about \( \tan(x/2) \).

### Step 1: Rewrite the given equation
We have:
\[ a^2 - 2a \cos x + 1 = 674 \]
This can be rearranged to:
\[ a^2 - 2a \cos x + 1 - 674 = 0 \]
or:
\[ a^2 - 2a \cos x - 673 = 0 \]

### Step 2: Use the identity for \( \cos x \)
We know that:
\[ \cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \]
Given \( \tan(x/2) = 7 \), we can substitute this into the identity:
\[ \cos x = \frac{1 - 7^2}{1 + 7^2} = \frac{1 - 49}{1 + 49} = \frac{-48}{50} = -\frac{24}{25} \]

### Step 3: Substitute \( \cos x \) back into the equation
Now substitute \( \cos x \) into the rearranged equation:
\[ a^2 - 2a \left(-\frac{24}{25}\right) - 673 = 0 \]
This simplifies to:
\[ a^2 + \frac{48a}{25} - 673 = 0 \]

### Step 4: Clear the fraction by multiplying through by 25
To eliminate the fraction, multiply the entire equation by 25:
\[ 25a^2 + 48a - 25 \times 673 = 0 \]
Calculating \( 25 \times 673 \):
\[ 25 \times 673 = 16825 \]
So the equation becomes:
\[ 25a^2 + 48a - 16825 = 0 \]

### Step 5: Apply the quadratic formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 25, b = 48, c = -16825 \):
\[ b^2 - 4ac = 48^2 - 4 \times 25 \times (-16825) \]
Calculating \( 48^2 \):
\[ 48^2 = 2304 \]
Calculating \( 4 \times 25 \times 16825 \):
\[ 4 \times 25 \times 16825 = 1682500 \]
So:
\[ b^2 - 4ac = 2304 + 1682500 = 1684804 \]

### Step 6: Calculate the square root
Now find \( \sqrt{1684804} \):
\[ \sqrt{1684804} = 1298 \]

### Step 7: Substitute back into the quadratic formula
Now substituting back:
\[ a = \frac{-48 \pm 1298}{50} \]
Calculating the two possible values:
1. \( a = \frac{-48 + 1298}{50} = \frac{1250}{50} = 25 \)
2. \( a = \frac{-48 - 1298}{50} = \frac{-1346}{50} = -26.92 \) (not an integer)

### Conclusion
The integral value of \( a \) is:
\[ \boxed{25} \]