Verify Rolles theorem for function `f(x)=sinx-sin2x` on `[0,\ pi]`

Explanation

To verify 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
We need to check if the function \( f(x) \) is continuous on the closed interval \([0, \pi]\).

- The function \( f(x) = \sin x - \sin 2x \) is composed of sine functions, which are continuous everywhere. Therefore, \( f(x) \) is continuous on \([0, \pi]\).

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

- The derivative of \( f(x) \) is given by:
\[ f'(x) = \cos x - 2\cos 2x \]
- Since \( \cos x \) and \( \cos 2x \) are also continuous and differentiable everywhere, \( f'(x) \) is differentiable on \((0, \pi)\).

### Step 3: Check Endpoints
Now we evaluate \( f(x) \) at the endpoints of the interval.

- Calculate \( f(0) \):
\[ f(0) = \sin(0) - \sin(2 \cdot 0) = 0 - 0 = 0 \]

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

### Step 4: Apply Rolle's Theorem
Since \( f(0) = f(\pi) \) and \( f(x) \) is continuous on \([0, \pi]\) and differentiable on \((0, \pi)\), we can apply Rolle's Theorem. According to Rolle's Theorem, there exists at least one \( c \in (0, \pi) \) such that \( f'(c) = 0 \).

### Step 5: Solve \( f'(c) = 0 \)
Set the derivative equal to zero:
\[ f'(c) = \cos c - 2\cos 2c = 0 \]

This implies:
\[ \cos c = 2\cos 2c \]

Using the double angle formula, we know:
\[ \cos 2c = 2\cos^2 c - 1 \]
Substituting this into the equation gives:
\[ \cos c = 2(2\cos^2 c - 1) \]
\[ \cos c = 4\cos^2 c - 2 \]
Rearranging this, we have:
\[ 4\cos^2 c - \cos c - 2 = 0 \]

### Step 6: Solve the Quadratic Equation
Let \( x = \cos c \). The equation becomes:
\[ 4x^2 - x - 2 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4} \]
\[ x = \frac{1 \pm \sqrt{1 + 32}}{8} = \frac{1 \pm \sqrt{33}}{8} \]

### Step 7: Verify \( c \) in the Interval
The solutions for \( \cos c \) are:
\[ \cos c = \frac{1 + \sqrt{33}}{8} \quad \text{and} \quad \cos c = \frac{1 - \sqrt{33}}{8} \]

We need to check if these values of \( c \) fall within the interval \( (0, \pi) \).

Since \( \cos c \) must be in the range \([-1, 1]\), we can check if \( \frac{1 + \sqrt{33}}{8} \) and \( \frac{1 - \sqrt{33}}{8} \) are valid.

### Conclusion
Since we have found at least one \( c \in (0, \pi) \) such that \( f'(c) = 0 \), we have verified Rolle's Theorem for the function \( f(x) = \sin x - \sin 2x \) on the interval \([0, \pi]\).

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