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 number of real solutions for \(x\).

### Step 1: Solve the first factor

The first factor is:

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

We can rewrite this as:

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

Let \(y = 2^{x}\), then \(2^{-x} = \frac{1}{y}\). This gives us:

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

Multiplying through by \(2y\) leads to:

\[ y^2 - 2 \cos x \cdot y + 1 = 0. \]

This is a quadratic equation in \(y\). The discriminant must be non-negative for real solutions:

\[ D = (2 \cos x)^2 - 4 \cdot 1 \cdot 1 = 4 \cos^2 x - 4 \geq 0. \]

Thus, we have:

\[ \cos^2 x \geq 1 \implies \cos x = \pm 1. \]

The solutions for \(\cos x = 1\) are:

\[ x = 2n\pi \quad (n \in \mathbb{Z}), \]

and for \(\cos x = -1\):

\[ x = (2n + 1)\pi \quad (n \in \mathbb{Z}). \]

### Step 2: Solve the second factor

The second factor is:

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

Rewriting gives:

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

Let \(z = 3^{x+\pi}\), then \(3^{-x-\pi} = \frac{1}{z}\). This leads to:

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

Multiplying through by \(2z\):

\[ z^2 + 2 \cos x \cdot z + 1 = 0. \]

The discriminant must be non-negative:

\[ D = (2 \cos x)^2 - 4 \cdot 1 \cdot 1 = 4 \cos^2 x - 4 \geq 0. \]

Thus, we have:

\[ \cos^2 x \geq 1 \implies \cos x = \pm 1. \]

This gives the same solutions as before.

### Step 3: Solve the third factor

The third factor is:

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

Rewriting gives:

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

Let \(w = 5^{\pi - x}\), then \(5^{x - \pi} = \frac{1}{w}\). This leads to:

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

Multiplying through by \(2w\):

\[ w^2 - 2 \cos x \cdot w + 1 = 0. \]

The discriminant must be non-negative:

\[ D = (2 \cos x)^2 - 4 \cdot 1 \cdot 1 = 4 \cos^2 x - 4 \geq 0. \]

Thus, we have:

\[ \cos^2 x \geq 1 \implies \cos x = \pm 1. \]

### Conclusion

From all three factors, we find that:

1. \(x = 2n\pi\) or \(x = (2n + 1)\pi\) from the first factor.
2. The same solutions from the second factor.
3. The same solutions from the third factor.

Thus, the number of distinct real solutions is:

- For \(n = 0\): \(x = 0\) (from \(2n\pi\)).
- For \(n = -1\): \(x = -\pi\) (from \((2n + 1)\pi\)).
- For \(n = 1\): \(x = 2\pi\) (from \(2n\pi\)).
- For \(n = -2\): \(x = -3\pi\) (from \((2n + 1)\pi\)).

The distinct solutions are \(0\), \(-\pi\), \(2\pi\), and \(-3\pi\).

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

\[ \boxed{4}. \]