If cos 4x = 1 + `k sin^2xcos^2x` then k =?

Options

A. 8
B. 4
C. -8
D. -4

Answer

To solve the equation \( \cos 4x = 1 + k \sin^2 x \cos^2 x \) and find the value of \( k \), we will use trigonometric identities.

### Step-by-step Solution:

1. **Use the Double Angle Formula for Cosine:**
We know that:
\[ \cos 4x = \cos(2 \cdot 2x) = 2\cos^2(2x) - 1 \]
Now, we need to express \( \cos(2x) \) in terms of \( \sin^2 x \) and \( \cos^2 x \).

2. **Apply the Double Angle Formula for Cosine Again:**
We can express \( \cos(2x) \) as:
\[ \cos(2x) = 2\cos^2 x - 1 \]
Substituting this into the equation for \( \cos 4x \):
\[ \cos 4x = 2(2\cos^2 x - 1)^2 - 1 \]

3. **Expand the Expression:**
First, calculate \( (2\cos^2 x - 1)^2 \):
\[ (2\cos^2 x - 1)^2 = 4\cos^4 x - 4\cos^2 x + 1 \]
Now substitute back:
\[ \cos 4x = 2(4\cos^4 x - 4\cos^2 x + 1) - 1 = 8\cos^4 x - 8\cos^2 x + 2 - 1 \]
Simplifying gives:
\[ \cos 4x = 8\cos^4 x - 8\cos^2 x + 1 \]

4. **Express \( \cos^2 x \) in terms of \( \sin^2 x \):**
We know that \( \cos^2 x = 1 - \sin^2 x \). Substitute this into the equation:
\[ \cos 4x = 8(1 - \sin^2 x)^2 - 8(1 - \sin^2 x) + 1 \]

5. **Expand and Simplify:**
Expanding \( (1 - \sin^2 x)^2 \):
\[ (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x \]
Therefore:
\[ \cos 4x = 8(1 - 2\sin^2 x + \sin^4 x) - 8(1 - \sin^2 x) + 1 \]
Simplifying this gives:
\[ \cos 4x = 8 - 16\sin^2 x + 8\sin^4 x - 8 + 8\sin^2 x + 1 \]
Combining like terms results in:
\[ \cos 4x = 8\sin^4 x - 8\sin^2 x + 1 \]

6. **Set the Equation Equal to the Given Form:**
Now we have:
\[ \cos 4x = 1 + k \sin^2 x \cos^2 x \]
We can express \( \sin^2 x \cos^2 x \) as \( \sin^2 x (1 - \sin^2 x) \):
\[ k \sin^2 x \cos^2 x = k \sin^2 x (1 - \sin^2 x) \]

7. **Compare Coefficients:**
From the equation \( 8\sin^4 x - 8\sin^2 x + 1 = 1 + k \sin^2 x (1 - \sin^2 x) \), we can compare coefficients:
- The coefficient of \( \sin^4 x \) gives \( k = -8 \).

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{-8} \]