Verify Rolles theorem for function `f(x)=2sinx+sin2x` on `[0,\ pi]`

Explanation

To verify Rolle's Theorem for the function \( f(x) = 2\sin x + \sin 2x \) 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 a function \( f \) is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

1. **Continuity**:
The function \( f(x) = 2\sin x + \sin 2x \) is a combination of sine functions, which are continuous everywhere. Therefore, \( f(x) \) is continuous on the interval \([0, \pi]\).
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