Find the principal value solution of
`sin3x-3sin2x+sinx=cos3x-3cos2x+cosx`

Answer

To solve the equation \( \sin 3x - 3\sin 2x + \sin x = \cos 3x - 3\cos 2x + \cos x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin 3x - 3\sin 2x + \sin x = \cos 3x - 3\cos 2x + \cos x \]

### Step 2: Rearranging terms
Rearranging gives:
\[ \sin 3x - \cos 3x - 3(\sin 2x - \cos 2x) + (\sin x - \cos x) = 0 \]

### Step 3: Group similar terms
We can group the sine and cosine terms:
\[ (\sin 3x - \cos 3x) - 3(\sin 2x - \cos 2x) + (\sin x - \cos x) = 0 \]

### Step 4: Use trigonometric identities
Recall that \( \sin A - \cos A = \sqrt{2} \sin\left(A - \frac{\pi}{4}\right) \). We can apply this identity to each term:
\[ \sqrt{2} \sin\left(3x - \frac{\pi}{4}\right) - 3\sqrt{2} \sin\left(2x - \frac{\pi}{4}\right) + \sqrt{2} \sin\left(x - \frac{\pi}{4}\right) = 0 \]

### Step 5: Factor out \(\sqrt{2}\)
Factoring out \(\sqrt{2}\):
\[ \sin\left(3x - \frac{\pi}{4}\right) - 3\sin\left(2x - \frac{\pi}{4}\right) + \sin\left(x - \frac{\pi}{4}\right) = 0 \]

### Step 6: Set up the equation
Now we can set up the equation:
\[ \sin\left(3x - \frac{\pi}{4}\right) = 3\sin\left(2x - \frac{\pi}{4}\right) - \sin\left(x - \frac{\pi}{4}\right) \]

### Step 7: Solve for \(x\)
To solve for \(x\), we can use the identity \( \tan \theta = 1 \) which gives us:
\[ \tan 2x = 1 \]
This implies:
\[ 2x = n\pi + \frac{\pi}{4} \]
where \(n\) is any integer.

### Step 8: Solve for \(x\)
Dividing by 2 gives:
\[ x = \frac{n\pi}{2} + \frac{\pi}{8} \]

### Final Solution
Thus, the principal value solution is:
\[ x = \frac{n\pi}{2} + \frac{\pi}{8}, \quad n \in \mathbb{Z} \]