The maximum values of `sinx(1+cosx)` will be at the

Options

A. `x=(pi)/(2)`
B. `x=(pi)/(6)`
C. `x=(pi)/(3)`
D. `x=pi`

Explanation

To find the maximum values of the function \( y = \sin x (1 + \cos x) \), we will follow these steps:

### Step 1: Differentiate the Function
We start by differentiating the function \( y \) with respect to \( x \).

\[ y = \sin x (1 + \cos x) \]

Using the product rule, we have:

\[ y' = \frac{d}{dx}(\sin x) \cdot (1 + \cos x) + \sin x \cdot \frac{d}{dx}(1 + \cos x) \]

Calculating the derivatives:

\[ y' = \cos x (1 + \cos x) + \sin x (-\sin x) \]

This simplifies to:

\[ y' = \cos x (1 + \cos x) - \sin^2 x \]

### Step 2: Set the Derivative to Zero
To find the critical points, we set \( y' = 0 \):

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

Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can rewrite the equation:

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

This simplifies to:

\[ \cos x + \cos^2 x - 1 + \cos^2 x = 0 \]
\[ 2\cos^2 x + \cos x - 1 = 0 \]

### Step 3: Solve the Quadratic Equation
Now, we solve the quadratic equation \( 2\cos^2 x + \cos x - 1 = 0 \) using the quadratic formula:

\[ \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]

This gives us two solutions:

1. \( \cos x = \frac{1}{2} \)
2. \( \cos x = -1 \)

### Step 4: Find the Corresponding Values of \( x \)
Now we find the values of \( x \):

1. For \( \cos x = \frac{1}{2} \):
- \( x = \frac{\pi}{3} + 2k\pi \) or \( x = -\frac{\pi}{3} + 2k\pi \) (where \( k \) is any integer)

2. For \( \cos x = -1 \):
- \( x = \pi + 2k\pi \)

### Step 5: Determine Maximum or Minimum
Next, we need to determine whether these critical points are maxima or minima by using the second derivative test.

### Step 6: Find the Second Derivative
We differentiate \( y' \) to find \( y'' \):

\[ y' = \cos x (1 + \cos x) - \sin^2 x \]

Differentiating again:

\[ y'' = -\sin x (1 + \cos x) - \sin x \cdot \cos x \]
\[ = -\sin x (1 + 2\cos x) \]

### Step 7: Evaluate the Second Derivative at Critical Points
1. For \( x = \frac{\pi}{3} \):
\[ y''\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right)(1 + 2\cos\left(\frac{\pi}{3}\right)) = -\frac{\sqrt{3}}{2}(1 + 2 \cdot \frac{1}{2}) = -\frac{\sqrt{3}}{2} \cdot 2 < 0 \]
This indicates a local maximum.

2. For \( x = \pi \):
\[ y''(\pi) = -\sin(\pi)(1 + 2\cos(\pi)) = 0 \]
This is inconclusive.

### Conclusion
The maximum value of \( y = \sin x (1 + \cos x) \) occurs at \( x = \frac{\pi}{3} \).