`f(x) = sinx - sin2x ` in `[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 the conditions of Rolle's Theorem

Rolle's Theorem states that if:
1. The function \( f(x) \) is continuous on the closed interval \([a, b]\).
2. The function \( f(x) \) is differentiable on the open interval \((a, b)\).
3. \( f(a) = f(b) \).

We will check these conditions for \( f(x) = \sin x - \sin 2x \) on the interval \([0, \pi]\).

### Step 2: Check continuity and differentiability

The function \( f(x) \) is a combination of sine functions, which are continuous and differentiable everywhere. Therefore, \( f(x) \) is continuous on \([0, \pi]\) and differentiable on \((0, \pi)\).

### Step 3: Evaluate \( f(0) \) and \( f(\pi) \)

Now we need to compute \( f(0) \) and \( f(\pi) \):

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

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

Since \( f(0) = f(\pi) = 0 \), the third condition \( f(a) = f(b) \) is satisfied.

### Step 4: Find the derivative \( f'(x) \)

Next, we find the derivative of \( f(x) \):

\[ f'(x) = \frac{d}{dx}(\sin x - \sin 2x) = \cos x - 2\cos 2x \]

### Step 5: Set the derivative equal to zero

To find the critical points, we set the derivative equal to zero:

\[ \cos x - 2\cos 2x = 0 \]

Using the double angle identity \( \cos 2x = 2\cos^2 x - 1 \), we can rewrite this as:

\[ \cos x - 2(2\cos^2 x - 1) = 0 \]

This simplifies to:

\[ \cos x - 4\cos^2 x + 2 = 0 \]

Rearranging gives us:

\[ 4\cos^2 x - \cos x - 2 = 0 \]

### Step 6: Solve the quadratic equation

Let \( u = \cos x \). The equation becomes:

\[ 4u^2 - u - 2 = 0 \]

Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ u = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4} \]

Calculating the discriminant:

\[ = \frac{1 \pm \sqrt{1 + 32}}{8} = \frac{1 \pm \sqrt{33}}{8} \]

### Step 7: Find the values of \( x \)

Thus, the solutions for \( u \) are:

\[ u = \frac{1 + \sqrt{33}}{8} \quad \text{and} \quad u = \frac{1 - \sqrt{33}}{8} \]

Now we need to check if these values are within the range of \( \cos x \) for \( x \in [0, \pi] \).

### Step 8: Verify the values of \( u \)

1. For \( u = \frac{1 + \sqrt{33}}{8} \), we check if it is in the range \([-1, 1]\).
2. For \( u = \frac{1 - \sqrt{33}}{8} \), we check if it is in the range \([-1, 1]\).

### Conclusion

Since both values of \( u \) yield valid \( x \) values in the interval \([0, \pi]\), we conclude that there exists at least one \( c \in (0, \pi) \) such that \( f'(c) = 0 \). Therefore, Rolle's Theorem is verified.

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