Suppose `sin^(3)x sin3x=sum_(m=0)^(n) C_(m) cos mx ` is an idedntity in x , where `C_(0),….C_(n)` are constant and `C_(n)ne0` then the value of n is ___________

Explanation

To solve the problem, we need to find the value of \( n \) in the identity:

\[ \sin^3 x \sin 3x = \sum_{m=0}^{n} C_m \cos mx \]

where \( C_0, C_1, \ldots, C_n \) are constants and \( C_n \neq 0 \).

### Step 1: Rewrite \( \sin 3x \)
We start by using the triple angle formula for sine:

\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]

### Step 2: Substitute \( \sin 3x \) into the equation
Now, substitute this expression into the left-hand side of the identity:

\[ \sin^3 x \sin 3x = \sin^3 x (3 \sin x - 4 \sin^3 x) \]

### Step 3: Distribute \( \sin^3 x \)
Distributing \( \sin^3 x \) gives us:

\[ = 3 \sin^4 x - 4 \sin^6 x \]

### Step 4: Express \( \sin^4 x \) and \( \sin^6 x \) in terms of cosine
We can express \( \sin^4 x \) and \( \sin^6 x \) using the identity \( \sin^2 x = 1 - \cos^2 x \):

1. For \( \sin^4 x \):

\[ \sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2 = 1 - 2\cos^2 x + \cos^4 x \]

2. For \( \sin^6 x \):

\[ \sin^6 x = (\sin^2 x)^3 = (1 - \cos^2 x)^3 = 1 - 3\cos^2 x + 3\cos^4 x - \cos^6 x \]

### Step 5: Substitute back into the equation
Substituting these into our expression:

\[ 3 \sin^4 x - 4 \sin^6 x = 3(1 - 2\cos^2 x + \cos^4 x) - 4(1 - 3\cos^2 x + 3\cos^4 x - \cos^6 x) \]

### Step 6: Simplify the expression
Now, simplify the expression:

\[ = 3 - 6\cos^2 x + 3\cos^4 x - 4 + 12\cos^2 x - 12\cos^4 x + 4\cos^6 x \]

Combining like terms:

\[ = -1 + 6\cos^2 x - 9\cos^4 x + 4\cos^6 x \]

### Step 7: Identify the highest power of cosine
The expression can be written as:

\[ = 4\cos^6 x - 9\cos^4 x + 6\cos^2 x - 1 \]

The highest power of \( \cos x \) in this expression is \( 6 \).

### Conclusion
Thus, the value of \( n \) is:

\[ \boxed{6} \]