If `cos4x=a_0+a_1cos^2x+a^2cos^4x` is true for all values of `x in R ,` then the value of `5a_0+a_1+a_2` is_______

Explanation

To solve the equation \( \cos 4x = a_0 + a_1 \cos^2 x + a_2 \cos^4 x \) for all values of \( x \in \mathbb{R} \), we will express \( \cos 4x \) in terms of \( \cos^2 x \).

### Step 1: Use the double angle formula for cosine
We start with the double angle formula:
\[ \cos 4x = 2 \cos^2 2x - 1 \]
Next, we need to express \( \cos 2x \) in terms of \( \cos^2 x \).

### Step 2: Express \( \cos 2x \) in terms of \( \cos^2 x \)
Using the double angle formula again:
\[ \cos 2x = 2 \cos^2 x - 1 \]
Now, substitute \( \cos 2x \) into the equation for \( \cos 4x \):
\[ \cos 4x = 2(2 \cos^2 x - 1)^2 - 1 \]

### Step 3: Expand \( (2 \cos^2 x - 1)^2 \)
Now we will expand \( (2 \cos^2 x - 1)^2 \):
\[ (2 \cos^2 x - 1)^2 = 4 \cos^4 x - 4 \cos^2 x + 1 \]
Substituting this back into the equation for \( \cos 4x \):
\[ \cos 4x = 2(4 \cos^4 x - 4 \cos^2 x + 1) - 1 \]

### Step 4: Simplify the equation
Now we simplify:
\[ \cos 4x = 8 \cos^4 x - 8 \cos^2 x + 2 - 1 \]
\[ \cos 4x = 8 \cos^4 x - 8 \cos^2 x + 1 \]

### Step 5: Compare coefficients
Now we compare this with the given equation:
\[ \cos 4x = a_0 + a_1 \cos^2 x + a_2 \cos^4 x \]
From the comparison, we can identify:
- \( a_2 = 8 \)
- \( a_1 = -8 \)
- \( a_0 = 1 \)

### Step 6: Calculate \( 5a_0 + a_1 + a_2 \)
Now we calculate \( 5a_0 + a_1 + a_2 \):
\[ 5a_0 + a_1 + a_2 = 5(1) + (-8) + 8 \]
\[ = 5 - 8 + 8 \]
\[ = 5 \]

Thus, the value of \( 5a_0 + a_1 + a_2 \) is \( \boxed{5} \).