The general solution of the equation `sin^4x + cos^4x= sinx cos x` is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the equation \( \sin^4 x + \cos^4 x = \sin x \cos x \), we can follow these steps:

### Step 1: Rewrite the left-hand side
We can use the identity \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \). Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]
Thus, the equation becomes:
\[ 1 - 2\sin^2 x \cos^2 x = \sin x \cos x \]

### Step 2: Rearranging the equation
Rearranging gives:
\[ 1 - \sin x \cos x - 2\sin^2 x \cos^2 x = 0 \]

### Step 3: Substitute \( \sin x \cos x \)
Let \( y = \sin x \cos x \). Then, we can express \( \sin^2 x \cos^2 x \) as \( y^2 \):
\[ 1 - y - 2y^2 = 0 \]

### Step 4: Rearranging into a standard quadratic form
Rearranging gives us:
\[ 2y^2 + y - 1 = 0 \]

### Step 5: Factor the quadratic equation
Now, we can factor this quadratic equation:
\[ (2y - 1)(y + 1) = 0 \]

### Step 6: Solve for \( y \)
Setting each factor to zero gives us:
1. \( 2y - 1 = 0 \) → \( y = \frac{1}{2} \)
2. \( y + 1 = 0 \) → \( y = -1 \)

### Step 7: Back-substituting for \( \sin x \cos x \)
Now we substitute back for \( y \):
1. \( \sin x \cos x = \frac{1}{2} \)
2. \( \sin x \cos x = -1 \)

### Step 8: Solve \( \sin x \cos x = \frac{1}{2} \)
Using the double angle identity \( \sin 2x = 2 \sin x \cos x \):
\[ \sin 2x = 1 \implies 2x = \frac{\pi}{2} + 2n\pi \implies x = \frac{\pi}{4} + n\pi \]

### Step 9: Solve \( \sin x \cos x = -1 \)
This case is not possible since \( \sin x \cos x \) cannot equal \(-1\) (the maximum value of \( \sin x \cos x \) is \( \frac{1}{2} \)).

### Final General Solution
Thus, the general solution to the equation \( \sin^4 x + \cos^4 x = \sin x \cos x \) is:
\[ x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]