If `sin(x +3alpha)=3 sin (alpha-x)`, then

Options

A. `tan x = tan alpha`
B. `tan x = tan^(2)alpha`
C. `tan x = tan^(3)alpha`
D. `tan x=3 tan alpha`

Explanation

To solve the equation \( \sin(x + 3\alpha) = 3 \sin(\alpha - x) \), we will use trigonometric identities and algebraic manipulation. Here is the step-by-step solution:

### Step 1: Apply the Sine Addition Formula
We start with the left side of the equation and apply the sine addition formula:
\[ \sin(x + 3\alpha) = \sin x \cos(3\alpha) + \cos x \sin(3\alpha) \]
Thus, we rewrite the equation as:
\[ \sin x \cos(3\alpha) + \cos x \sin(3\alpha) = 3 \sin(\alpha - x) \]

### Step 2: Apply the Sine Subtraction Formula
Now, we apply the sine subtraction formula on the right side:
\[ \sin(\alpha - x) = \sin \alpha \cos x - \cos \alpha \sin x \]
So, we can rewrite the equation as:
\[ \sin x \cos(3\alpha) + \cos x \sin(3\alpha) = 3(\sin \alpha \cos x - \cos \alpha \sin x) \]

### Step 3: Rearranging the Equation
Rearranging gives us:
\[ \sin x \cos(3\alpha) + \cos x \sin(3\alpha) = 3 \sin \alpha \cos x - 3 \cos \alpha \sin x \]
This can be rearranged to:
\[ \sin x \cos(3\alpha) + 3 \cos \alpha \sin x = 3 \sin \alpha \cos x - \cos x \sin(3\alpha) \]

### Step 4: Grouping Terms
Now, we can group the terms involving \(\sin x\) and \(\cos x\):
\[ \sin x (\cos(3\alpha) + 3 \cos \alpha) = \cos x (3 \sin \alpha - \sin(3\alpha)) \]

### Step 5: Factor Out Common Terms
We can factor out \(\sin x\) and \(\cos x\):
\[ \frac{\sin x}{\cos x} = \frac{3 \sin \alpha - \sin(3\alpha)}{\cos(3\alpha) + 3 \cos \alpha} \]
This simplifies to:
\[ \tan x = \frac{3 \sin \alpha - \sin(3\alpha)}{\cos(3\alpha) + 3 \cos \alpha} \]

### Step 6: Use the Sine and Cosine Triple Angle Formulas
Using the triple angle formulas:
\[ \sin(3\alpha) = 3 \sin \alpha - 4 \sin^3 \alpha \]
\[ \cos(3\alpha) = 4 \cos^3 \alpha - 3 \cos \alpha \]
Substituting these into the equation gives:
\[ \tan x = \frac{3 \sin \alpha - (3 \sin \alpha - 4 \sin^3 \alpha)}{(4 \cos^3 \alpha - 3 \cos \alpha) + 3 \cos \alpha} \]
This simplifies to:
\[ \tan x = \frac{4 \sin^3 \alpha}{4 \cos^3 \alpha} \]
Thus:
\[ \tan x = \tan^3 \alpha \]

### Step 7: Final Result
From this, we conclude:
\[ \tan x = \tan^3 \alpha \]
This implies:
\[ \tan x = \tan^3 \alpha \implies 10 x = 10^3 \alpha \]

### Conclusion
Thus, the final answer is:
\[ 10 x = 10^3 \alpha \]