Let `f(x)=cos10x+cos8x+3cos4x+3cos2x` and `g(x)=8cosxcos^3(3x)` then for all `x ` we have

Options

A. `f(x) = g(x)`
B. `2f(x) = 3g(x)`
C. `f(x) = 2g(x)`
D. `2f(x) = g(x)`

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given by:

\[ f(x) = \cos 10x + \cos 8x + 3\cos 4x + 3\cos 2x \]
\[ g(x) = 8\cos x \cos^3(3x) \]

### Step 1: Simplifying \( f(x) \)

We will start by simplifying \( f(x) \). We can group the terms in pairs and apply the cosine addition formula:

\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

#### Pairing the terms:

1. Pair \( \cos 10x \) and \( \cos 8x \):
- Let \( A = 10x \) and \( B = 8x \)
- Then, we have:
\[ \cos 10x + \cos 8x = 2 \cos\left(\frac{10x + 8x}{2}\right) \cos\left(\frac{10x - 8x}{2}\right) = 2 \cos(9x) \cos(x) \]

2. Pair \( 3\cos 4x \) and \( 3\cos 2x \):
- Factor out the 3:
\[ 3(\cos 4x + \cos 2x) = 3 \cdot 2 \cos\left(\frac{4x + 2x}{2}\right) \cos\left(\frac{4x - 2x}{2}\right) = 6 \cos(3x) \cos(x) \]

#### Combining the results:

Now we can combine the results from the two pairs:
\[ f(x) = 2 \cos(9x) \cos(x) + 6 \cos(3x) \cos(x) \]
Factoring out \( \cos(x) \):
\[ f(x) = \cos(x) \left(2 \cos(9x) + 6 \cos(3x)\right) \]

### Step 2: Further simplification of \( f(x) \)

Using the identity \( \cos(3x) = 4\cos^3(x) - 3\cos(x) \):
\[ f(x) = \cos(x) \left(2 \cos(9x) + 6(4\cos^3(x) - 3\cos(x))\right) \]
\[ = \cos(x) \left(2 \cos(9x) + 24\cos^3(x) - 18\cos(x)\right) \]

### Step 3: Relating \( f(x) \) to \( g(x) \)

Now, we will express \( g(x) \):
\[ g(x) = 8 \cos(x) \cos^3(3x) \]
Using the identity for \( \cos(3x) \):
\[ g(x) = 8 \cos(x) (4\cos^3(x) - 3\cos(x))^3 \]
This simplifies to:
\[ g(x) = 8 \cos(x) \cdot 4\cos^3(3x) \]

### Step 4: Conclusion

Now we see that:
\[ f(x) = 2 \cos(x) (4 \cos^3(3x) - 3 \cos(3x)) + 6 \cos(3x) \cos(x) \]
After simplification, we find that:
\[ f(x) = g(x) \]

Thus, we conclude that:
\[ f(x) = g(x) \]

### Final Answer:
\[ f(x) = g(x) \quad \text{for all } x \]

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