If `a sin x+b cos(c+x)+b cos(c-x)=alpha, alpha gt a`, then the minimum value of `|cos c|` is :

Options

A. `sqrt((alpha^(2)-a^(2))/(b^(2)))`
B. `sqrt((alpha^(2)-a^(2))/(2b^(2)))`
C. `sqrt((alpha^(2)-a^(2))/(3b^(2)))`
D. `sqrt((alpha^(2)-a^(2))/(4b^(2)))`

Answer

To solve the problem, we need to find the minimum value of \( | \cos c | \) given the equation:

\[ a \sin x + b \cos(c + x) + b \cos(c - x) = \alpha \]

where \( \alpha > a \).

### Step-by-Step Solution:

1. **Rewrite the equation**:
\[ a \sin x + b \left( \cos(c + x) + \cos(c - x) \right) = \alpha \]

2. **Use the cosine addition formula**:
The sum of cosines can be simplified using the formula:
\[ \cos(c + x) + \cos(c - x) = 2 \cos c \cos x \]
Thus, the equation becomes:
\[ a \sin x + 2b \cos c \cos x = \alpha \]

3. **Rearranging the equation**:
We can rearrange this to isolate \( \cos c \):
\[ 2b \cos c \cos x = \alpha - a \sin x \]
Therefore,
\[ \cos c = \frac{\alpha - a \sin x}{2b \cos x} \]

4. **Finding the minimum value**:
To find the minimum value of \( | \cos c | \), we need to analyze the expression:
\[ | \cos c | = \left| \frac{\alpha - a \sin x}{2b \cos x} \right| \]

5. **Using the identity for sine and cosine**:
We know that \( \sin^2 x + \cos^2 x = 1 \). Thus, we can express \( \cos^2 x \) as \( 1 - \sin^2 x \).

6. **Substituting \( \sin x \)**:
Let \( \sin x = k \) where \( -1 \leq k \leq 1 \). Then \( \cos x = \sqrt{1 - k^2} \). The expression for \( | \cos c | \) becomes:
\[ | \cos c | = \left| \frac{\alpha - ak}{2b \sqrt{1 - k^2}} \right| \]

7. **Differentiating to find critical points**:
To find the minimum, we can differentiate the expression with respect to \( k \) and set the derivative to zero. However, we can also analyze the endpoints and critical points directly.

8. **Setting \( \sin x = \frac{a}{\alpha} \)**:
From the condition \( \alpha > a \), we can find \( k \) such that:
\[ \sin x = \frac{a}{\alpha} \]
This gives us:
\[ \cos x = \sqrt{1 - \left(\frac{a}{\alpha}\right)^2} = \frac{\sqrt{\alpha^2 - a^2}}{\alpha} \]

9. **Substituting back**:
Substituting \( \sin x \) back into the equation for \( | \cos c | \):
\[ | \cos c | = \left| \frac{\alpha - a \cdot \frac{a}{\alpha}}{2b \cdot \frac{\sqrt{\alpha^2 - a^2}}{\alpha}} \right| \]
Simplifying this gives:
\[ | \cos c | = \frac{\alpha^2 - a^2}{2b \sqrt{\alpha^2 - a^2}} \]

10. **Final expression**:
Therefore, the minimum value of \( | \cos c | \) is:
\[ | \cos c | = \frac{\sqrt{\alpha^2 - a^2}}{2b} \]

### Conclusion:
The minimum value of \( | \cos c | \) is:

\[ \frac{\sqrt{\alpha^2 - a^2}}{2b} \]