If `sin^3 x + sin x cos x + cos^3 x = 1`, then x =

Options

A. `2n pi`
B. `2n pi + pi/2 `
C. `2n pi-pi/2`
D. `n pi`

Answer

To solve the equation \( \sin^3 x + \sin x \cos x + \cos^3 x = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We know that \( \sin^3 x + \cos^3 x \) can be factored using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, let \( a = \sin x \) and \( b = \cos x \).

Thus, we can rewrite the equation as:
\[ \sin^3 x + \cos^3 x + \sin x \cos x = 1 \]

### Step 2: Apply the identity
Using the identity, we have:
\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we can simplify this to:
\[ \sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x) \]

### Step 3: Substitute back into the equation
Substituting this back into our equation gives:
\[ (\sin x + \cos x)(1 - \sin x \cos x) + \sin x \cos x = 1 \]
This simplifies to:
\[ (\sin x + \cos x)(1 - \sin x \cos x) + \sin x \cos x = 1 \]
\[ (\sin x + \cos x) - (\sin x + \cos x) \sin x \cos x + \sin x \cos x = 1 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \sin x + \cos x - \sin x \cos x (\sin x + \cos x) = 1 \]
Let \( S = \sin x + \cos x \), then:
\[ S - S \sin x \cos x = 1 \]
Factoring out \( S \):
\[ S(1 - \sin x \cos x) = 1 \]

### Step 5: Solve for \( \sin x \cos x \)
From the equation \( S(1 - \sin x \cos x) = 1 \), we can express \( \sin x \cos x \):
\[ 1 - \sin x \cos x = \frac{1}{S} \]
Thus,
\[ \sin x \cos x = 1 - \frac{1}{S} \]

### Step 6: Use the identity for \( S \)
Recall that \( S = \sin x + \cos x \) can be expressed in terms of \( \sin 2x \):
\[ S^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x \]
Let \( \sin x \cos x = t \):
\[ S^2 = 1 + 2t \]

### Step 7: Substitute back and solve
Substituting \( t \) back into our equation gives:
\[ S(1 - t) = 1 \]
This leads to a quadratic in terms of \( S \) and \( t \).

### Step 8: Solve for \( x \)
After solving the equations, we find:
1. \( \sin x + \cos x = 1 \) leads to \( x = \frac{\pi}{4} + n\pi \).
2. \( \sin x \cos x = 0 \) leads to \( x = n\frac{\pi}{2} \).

### Final Answer
Thus, the solutions for \( x \) are:
\[ x = \frac{\pi}{4} + n\pi \quad \text{and} \quad x = n\frac{\pi}{2} \]