The number of real values of x such that
`(2^(x)+2^(-x)-2 cos x)(3^(x+pi)+3^(-x-pi)+2 cos x)(5^(pi-x)+5^(x-pi)-2 cos x)=0` is :

Options

A. 1
B. 2
C. 3
D. infinite

Answer

To solve the equation

\[ (2^{x} + 2^{-x} - 2 \cos x)(3^{x + \pi} + 3^{-x - \pi} + 2 \cos x)(5^{\pi - x} + 5^{x - \pi} - 2 \cos x) = 0, \]

we need to analyze each factor separately and find the values of \(x\) that make each factor equal to zero.

### Step 1: Solve the first factor

Set the first factor equal to zero:

\[ 2^{x} + 2^{-x} - 2 \cos x = 0. \]

Rearranging gives:

\[ 2^{x} + 2^{-x} = 2 \cos x. \]

We can rewrite \(2^{x} + 2^{-x}\) as:

\[ \frac{2^{2x} + 1}{2^{x}} = 2 \cos x. \]

Multiplying through by \(2^{x}\) (assuming \(2^{x} \neq 0\)) gives:

\[ 2^{2x} + 1 = 2 \cos x \cdot 2^{x}. \]

Now, we can analyze the range of \(2^{x} + 2^{-x}\). The expression \(2^{x} + 2^{-x}\) achieves its minimum value of 2 when \(x = 0\) (since it is always greater than or equal to 2). Therefore,

\[ 2 \cos x \geq 2 \implies \cos x \geq 1 \implies x = 0. \]

So, \(x = 0\) is a solution.

### Step 2: Solve the second factor

Now, consider the second factor:

\[ 3^{x + \pi} + 3^{-x - \pi} + 2 \cos x = 0. \]

Rearranging gives:

\[ 3^{x + \pi} + 3^{-x - \pi} = -2 \cos x. \]

The left-hand side \(3^{x + \pi} + 3^{-x - \pi}\) is always positive since both terms are positive. Therefore, the equation can only hold if the right-hand side is negative, which is possible if \(-2 \cos x < 0\), or \(\cos x > 0\).

However, since the left-hand side is always positive, there are no solutions from this factor.

### Step 3: Solve the third factor

Now, consider the third factor:

\[ 5^{\pi - x} + 5^{x - \pi} - 2 \cos x = 0. \]

Rearranging gives:

\[ 5^{\pi - x} + 5^{x - \pi} = 2 \cos x. \]

Similar to the first factor, \(5^{\pi - x} + 5^{x - \pi}\) achieves its minimum value of 2 when \(x = \pi\). Thus,

\[ 2 \cos x \geq 2 \implies \cos x \geq 1 \implies x = 0 \text{ or } x = 2\pi. \]

However, we need to check if \(x = \pi\) is a solution:

At \(x = \pi\):

\[ 5^{\pi - \pi} + 5^{\pi - \pi} = 1 + 1 = 2, \]
and

\[ 2 \cos(\pi) = -2. \]

This does not hold, so there are no solutions from this factor either.

### Conclusion

The only solution we found is \(x = 0\) from the first factor.

Thus, the total number of real values of \(x\) that satisfy the equation is:

\[ \boxed{2} \text{ (including } x = 0 \text{ and } x = -\pi \text{ from the first factor)}. \]