Verify the conditions of Rolle's Theorem in the following problems. In each case, find a point in the interval, where the derivative vanishes :
`sinx-sin2x" on "[0,pi]`

Answer

To verify the conditions of Rolle's Theorem for the function \( f(x) = \sin x - \sin 2x \) on the interval \([0, \pi]\), we will follow these steps:

### Step 1: Check Continuity
Rolle's Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and if \(f(a) = f(b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\).

1. **Function Definition**: Let \( f(x) = \sin x - \sin 2x \).
2. **Check Continuity**: The sine function is continuous for all \(x\). Therefore, \(f(x)\) is continuous on the interval \([0, \pi]\).

### Step 2: Check Differentiability
Next, we need to check if the function is differentiable on the open interval \((0, \pi)\).

1. **Differentiate the Function**:
\[ f'(x) = \frac{d}{dx}(\sin x - \sin 2x) = \cos x - 2\cos 2x \]

2. **Check Differentiability**: The derivative \(f'(x)\) is composed of continuous functions (cosine functions), hence \(f'(x)\) is differentiable on \((0, \pi)\).

### Step 3: Check Endpoint Values
Now we need to check if \(f(0) = f(\pi)\).

1. **Calculate \(f(0)\)**:
\[ f(0) = \sin(0) - \sin(0) = 0 - 0 = 0 \]

2. **Calculate \(f(\pi)\)**:
\[ f(\pi) = \sin(\pi) - \sin(2\pi) = 0 - 0 = 0 \]

Since \(f(0) = f(\pi)\), we have verified that the third condition of Rolle's Theorem is satisfied.

### Step 4: Find \(c\) such that \(f'(c) = 0\)
Now we need to find a point \(c\) in the interval \((0, \pi)\) where \(f'(c) = 0\).

1. **Set the Derivative to Zero**:
\[ f'(x) = \cos x - 2\cos 2x = 0 \]

2. **Rearranging the Equation**:
\[ \cos x = 2\cos 2x \]

3. **Using the Double Angle Formula**:
Recall that \(\cos 2x = 2\cos^2 x - 1\), so we can substitute:
\[ \cos x = 2(2\cos^2 x - 1) = 4\cos^2 x - 2 \]

4. **Rearranging**:
\[ 4\cos^2 x - \cos x - 2 = 0 \]

5. **Applying the Quadratic Formula**:
Let \(u = \cos x\). The equation becomes:
\[ 4u^2 - u - 2 = 0 \]
Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ u = \frac{1 \pm \sqrt{1 + 32}}{8} = \frac{1 \pm \sqrt{33}}{8} \]

6. **Finding \(c\)**:
Since \(u = \cos c\), we have:
\[ c = \cos^{-1}\left(\frac{1 + \sqrt{33}}{8}\right) \quad \text{or} \quad c = \cos^{-1}\left(\frac{1 - \sqrt{33}}{8}\right) \]

### Conclusion
Thus, we have verified that all conditions of Rolle's Theorem are satisfied, and we found the points \(c\) where the derivative vanishes.