Let f(x) = sin x (1+cos x) ,` x in (0,2pi).` Find the number of critical points of f(x) . Also identify which of these critical points are points of Maxima`//`Minima.

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \sin x (1 + \cos x) \), where \( x \in (0, 2\pi) \).

### Step 2: Find the first derivative
We will use the product rule to differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}[\sin x] \cdot (1 + \cos x) + \sin x \cdot \frac{d}{dx}[1 + \cos x] \]
Calculating the derivatives:
\[ \frac{d}{dx}[\sin x] = \cos x \]
\[ \frac{d}{dx}[1 + \cos x] = -\sin x \]
Thus,
\[ f'(x) = \cos x (1 + \cos x) + \sin x (-\sin x) \]
\[ = \cos x (1 + \cos x) - \sin^2 x \]
Using the identity \( \sin^2 x = 1 - \cos^2 x \):
\[ f'(x) = \cos x + \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x + \cos x - 1 \]

### Step 3: Set the first derivative to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ 2\cos^2 x + \cos x - 1 = 0 \]
Let \( u = \cos x \). The equation becomes:
\[ 2u^2 + u - 1 = 0 \]
Using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us two solutions:
\[ u = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad u = \frac{-4}{4} = -1 \]
Thus, we have:
1. \( \cos x = \frac{1}{2} \)
2. \( \cos x = -1 \)

### Step 4: Find the values of \( x \)
For \( \cos x = \frac{1}{2} \):
\[ x = \frac{\pi}{3}, \frac{5\pi}{3} \]
For \( \cos x = -1 \):
\[ x = \pi \]

### Step 5: List the critical points
The critical points in the interval \( (0, 2\pi) \) are:
\[ x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \]

### Step 6: Determine maxima and minima using the second derivative
Now we find the second derivative \( f''(x) \):
\[ f''(x) = \frac{d}{dx}[f'(x)] = \frac{d}{dx}[2\cos^2 x + \cos x - 1] \]
Using the chain rule:
\[ f''(x) = 2 \cdot 2\cos x (-\sin x) + (-\sin x) = -4\cos x \sin x - \sin x = -\sin x (4\cos x + 1) \]

Now we evaluate \( f''(x) \) at the critical points:
1. At \( x = \frac{\pi}{3} \):
\[ f''\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right)(4\cos\left(\frac{\pi}{3}\right) + 1) = -\frac{\sqrt{3}}{2}(4 \cdot \frac{1}{2} + 1) = -\frac{\sqrt{3}}{2}(2 + 1) = -\frac{3\sqrt{3}}{2} < 0 \quad \text{(Maxima)} \]

2. At \( x = \pi \):
\[ f''(\pi) = -\sin(\pi)(4\cos(\pi) + 1) = -0 \quad \text{(Inconclusive)} \]

3. At \( x = \frac{5\pi}{3} \):
\[ f''\left(\frac{5\pi}{3}\right) = -\sin\left(\frac{5\pi}{3}\right)(4\cos\left(\frac{5\pi}{3}\right) + 1) = -\left(-\frac{\sqrt{3}}{2}\right)(4 \cdot \frac{1}{2} + 1) = \frac{\sqrt{3}}{2}(2 + 1) = \frac{3\sqrt{3}}{2} > 0 \quad \text{(Minima)} \]

### Conclusion
- The critical points are \( x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \).
- \( x = \frac{\pi}{3} \) is a point of maxima.
- \( x = \frac{5\pi}{3} \) is a point of minima.
- \( x = \pi \) is inconclusive.