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

### Step 1: Define the function
Let \( y = \sin x (1 + \cos x) \).

### Step 2: Differentiate the function
To find the maximum value, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} (\sin x (1 + \cos x)) \]
Using the product rule:
\[ \frac{dy}{dx} = \cos x (1 + \cos x) + \sin x (-\sin x) \]
This simplifies to:
\[ \frac{dy}{dx} = \cos x (1 + \cos x) - \sin^2 x \]

### Step 3: Set the derivative to zero
To find critical points, set \( \frac{dy}{dx} = 0 \):
\[ \cos x (1 + \cos x) - \sin^2 x = 0 \]
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 4: Solve the quadratic equation
Let \( u = \cos x \). The equation becomes:
\[ 2u^2 + u - 1 = 0 \]
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ u = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives:
\[ u = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad u = \frac{-4}{4} = -1 \]

### Step 5: Find the corresponding values of \( x \)
1. For \( \cos x = \frac{1}{2} \):
\[ x = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]
2. For \( \cos x = -1 \):
\[ x = \pi + 2n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Determine maximum or minimum
To determine whether these points are maxima or minima, we can use the second derivative test. We find \( \frac{d^2y}{dx^2} \) and evaluate it at \( x = \frac{\pi}{3} \).

### Step 7: Calculate the second derivative
Using the first derivative:
\[ \frac{dy}{dx} = \cos x (1 + \cos x) - \sin^2 x \]
Differentiate again:
\[ \frac{d^2y}{dx^2} = -\sin x (1 + \cos x) - 2\cos x \sin x \]
At \( x = \frac{\pi}{3} \):
\[ \frac{d^2y}{dx^2} = -\sin\left(\frac{\pi}{3}\right)(1 + \cos\left(\frac{\pi}{3}\right)) - 2\cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{3}\right) \]
Calculating this gives a negative value, confirming that \( x = \frac{\pi}{3} \) is a maximum.

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

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