Which of the following is not the solution of the equation `sin 5x=16 sin^(5) x(n inZ)` ?

Options

A. `n pi`
B. `npi+pi/6`
C. `npi- pi/6`
D. none of these

Explanation

To solve the equation \( \sin 5x = 16 \sin^5 x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin 5x = 16 \sin^5 x \]

### Step 2: Use the angle addition formula
We can express \( \sin 5x \) using the angle addition formula. We can rewrite \( 5x \) as \( 3x + 2x \):
\[ \sin 5x = \sin(3x + 2x) = \sin 3x \cos 2x + \cos 3x \sin 2x \]

However, for this problem, we will simplify it differently.

### Step 3: Factor the equation
Rearranging gives:
\[ \sin 5x - 16 \sin^5 x = 0 \]
This can be factored as:
\[ \sin 5x - 16 \sin^5 x = 0 \implies \sin 5x = 16 \sin^5 x \]

### Step 4: Use the identity for \( \sin 5x \)
Using the identity for \( \sin 5x \):
\[ \sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x \]
We can substitute this into our equation:
\[ 5 \sin x - 20 \sin^3 x + 16 \sin^5 x = 16 \sin^5 x \]

### Step 5: Simplify the equation
Subtract \( 16 \sin^5 x \) from both sides:
\[ 5 \sin x - 20 \sin^3 x = 0 \]

### Step 6: Factor out common terms
Factoring out \( \sin x \):
\[ \sin x (5 - 20 \sin^2 x) = 0 \]

### Step 7: Solve for \( \sin x = 0 \)
Setting \( \sin x = 0 \):
\[ x = n\pi, \quad n \in \mathbb{Z} \]

### Step 8: Solve for \( 5 - 20 \sin^2 x = 0 \)
Setting \( 5 - 20 \sin^2 x = 0 \):
\[ 20 \sin^2 x = 5 \implies \sin^2 x = \frac{1}{4} \]
Thus:
\[ \sin x = \pm \frac{1}{2} \]

### Step 9: Find solutions for \( \sin x = \frac{1}{2} \) and \( \sin x = -\frac{1}{2} \)
For \( \sin x = \frac{1}{2} \):
\[ x = n\pi + \frac{\pi}{6}, \quad n \in \mathbb{Z} \]
For \( \sin x = -\frac{1}{2} \):
\[ x = n\pi - \frac{\pi}{6}, \quad n \in \mathbb{Z} \]

### Summary of Solutions
The solutions to the equation \( \sin 5x = 16 \sin^5 x \) are:
1. \( x = n\pi \)
2. \( x = n\pi + \frac{\pi}{6} \)
3. \( x = n\pi - \frac{\pi}{6} \)

### Conclusion
Now, we need to identify which of the given options is not a solution.