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
We are given the function:
\[ f(x) = \sin x (1 + \cos x) \]

### Step 2: Find the first derivative
To find the critical points, we need to find the first derivative \( f'(x) \). We will use the product rule for differentiation:
\[ f'(x) = \frac{d}{dx}[\sin x] \cdot (1 + \cos x) + \sin x \cdot \frac{d}{dx}[1 + \cos x] \]

Calculating the derivatives:
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( 1 + \cos x \) is \( -\sin x \).

Thus, we have:
\[ f'(x) = \cos x (1 + \cos x) + \sin x (-\sin x) \]
\[ = \cos x (1 + \cos x) - \sin^2 x \]
\[ = \cos x + \cos^2 x - \sin^2 x \]

Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite:
\[ \sin^2 x = 1 - \cos^2 x \]
So,
\[ f'(x) = \cos x + \cos^2 x - (1 - \cos^2 x) \]
\[ = \cos x + 2\cos^2 x - 1 \]

### Step 3: Set the first derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 2\cos^2 x + \cos x - 1 = 0 \]

This is a quadratic equation in terms of \( \cos x \). Let \( u = \cos x \):
\[ 2u^2 + u - 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ = \frac{-1 \pm \sqrt{1 + 8}}{4} \]
\[ = \frac{-1 \pm 3}{4} \]

Calculating the roots:
1. \( u = \frac{2}{4} = \frac{1}{2} \)
2. \( u = \frac{-4}{4} = -1 \)

### Step 5: Find the values of \( x \)
Now we find the angles corresponding to these values of \( \cos x \):
1. \( \cos x = \frac{1}{2} \) gives \( x = \frac{\pi}{3}, \frac{5\pi}{3} \)
2. \( \cos x = -1 \) gives \( x = \pi \)

Thus, the critical points are:
\[ x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \]

### Step 6: Determine maxima and minima
To classify these critical points, we need to find the second derivative \( f''(x) \).

### Step 7: Find the second derivative
We differentiate \( f'(x) = 2\cos^2 x + \cos x - 1 \):
\[ f''(x) = \frac{d}{dx}[2\cos^2 x] + \frac{d}{dx}[\cos x] = 2 \cdot 2\cos x(-\sin x) - \sin x \]
\[ = -4\cos x \sin x - \sin x \]
\[ = -\sin x (4\cos x + 1) \]

### Step 8: Evaluate the second derivative at critical points
1. At \( x = \frac{\pi}{3} \):
\[ f''\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right)(4\cos\left(\frac{\pi}{3}\right) + 1) = -\frac{\sqrt{3}}{2}(4 \cdot \frac{1}{2} + 1) = -\frac{\sqrt{3}}{2}(2 + 1) = -\frac{3\sqrt{3}}{2} < 0 \]
(Maxima)

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

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

### Final Result
- Critical points: \( x = \frac{\pi}{3}, \pi, \frac{5\pi}{3} \)
- Maxima at \( x = \frac{\pi}{3} \)
- Minima at \( x = \frac{5\pi}{3} \)