If `sin^(3)theta+cos^(3)theta=0`, then what is the value of `theta` ?

Options

A. `(-pi)/(4)`
B. 0
C. `(pi)/(4)`
D. `(pi)/(3)`

Explanation

To solve the equation \( \sin^3 \theta + \cos^3 \theta = 0 \), we can follow these steps:

### Step 1: Apply the Sum of Cubes Identity
We know that \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, let \( a = \sin \theta \) and \( b = \cos \theta \). Thus, we can rewrite the equation as:

\[ \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) = 0 \]

### Step 2: Set Each Factor to Zero
For the product to be zero, either factor must be zero. Therefore, we have two cases to consider:

1. \( \sin \theta + \cos \theta = 0 \)
2. \( \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta = 0 \)

### Step 3: Solve the First Case
From the first case, \( \sin \theta + \cos \theta = 0 \), we can rearrange this to:

\[ \sin \theta = -\cos \theta \]

Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)), we get:

\[ \tan \theta = -1 \]

### Step 4: Find Values of \( \theta \)
The general solution for \( \tan \theta = -1 \) is:

\[ \theta = -\frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 5: Solve the Second Case
For the second case, we have:

\[ \sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta = 0 \]

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute:

\[ 1 - \sin \theta \cos \theta = 0 \]

This implies:

\[ \sin \theta \cos \theta = 1 \]

However, since \( \sin \theta \) and \( \cos \theta \) can never simultaneously equal 1 (the maximum value for both is 1), this case does not yield any valid solutions.

### Conclusion
Thus, the only valid solutions come from the first case:

\[ \theta = -\frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
The value of \( \theta \) is \( -\frac{\pi}{4} + n\pi \).

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