If `a sin x+b cos(x+theta)+b cos(x-theta)=d,` then the minimum value of `|cos theta|` is equal to

Options

A. `(1)/(2|b|)sqrt(d^(2)-a^(2))`
B. `(1)/(2|a|)sqrt(d^(2)-a^(2))`
C. `(1)/(2|d|)sqrt(d^(2)-a^(2))`
D. none of these.

Explanation

To solve the equation \( a \sin x + b \cos(x + \theta) + b \cos(x - \theta) = d \), we will follow these steps:

### Step 1: Rewrite the cosine terms
Using the cosine addition formula, we can rewrite \( \cos(x + \theta) \) and \( \cos(x - \theta) \):
\[ \cos(x + \theta) = \cos x \cos \theta - \sin x \sin \theta \]
\[ \cos(x - \theta) = \cos x \cos \theta + \sin x \sin \theta \]
Substituting these into the original equation gives:
\[ a \sin x + b (\cos x \cos \theta - \sin x \sin \theta) + b (\cos x \cos \theta + \sin x \sin \theta) = d \]

### Step 2: Combine like terms
Combining the terms results in:
\[ a \sin x + b \cos x \cos \theta - b \sin x \sin \theta + b \cos x \cos \theta + b \sin x \sin \theta = d \]
This simplifies to:
\[ a \sin x + 2b \cos x \cos \theta = d \]

### Step 3: Rearranging the equation
Rearranging gives us:
\[ a \sin x + 2b \cos x \cos \theta - d = 0 \]

### Step 4: Identify the quadratic form
This equation can be viewed as a quadratic in terms of \( \tan(x) \). By substituting \( \sin x = \frac{2\tan(x/2)}{1 + \tan^2(x/2)} \) and \( \cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \), we can express the equation in terms of \( \tan^2(x/2) \).

### Step 5: Discriminant condition
For \( \tan^2(x/2) \) to have real solutions, the discriminant must be non-negative:
\[ B^2 - 4AC \geq 0 \]
Here, \( A = -2A \), \( B = 2b \cos \theta \), and \( C = d - 2b \cos \theta \).

### Step 6: Set up the discriminant
The discriminant becomes:
\[ (2b \cos \theta)^2 - 4(-2A)(d - 2b \cos \theta) \geq 0 \]
This simplifies to:
\[ 4b^2 \cos^2 \theta + 8A(d - 2b \cos \theta) \geq 0 \]

### Step 7: Solve for \( \cos^2 \theta \)
Rearranging gives:
\[ b^2 \cos^2 \theta \geq \frac{d^2 - A^2}{4} \]
Thus:
\[ \cos^2 \theta \geq \frac{d^2 - A^2}{4b^2} \]

### Step 8: Find the minimum value of \( | \cos \theta | \)
Taking the square root gives:
\[ | \cos \theta | \geq \frac{\sqrt{d^2 - A^2}}{2b} \]

### Conclusion
The minimum value of \( | \cos \theta | \) is:
\[ \frac{\sqrt{d^2 - A^2}}{2b} \]