The equation `sin x + x cos x = 0` has at least one root in

Options

A. `(-pi//2,0)`
B. `(0,pi)`
C. `(-pi//2,pi//2)`
D. none of these

Explanation

To solve the equation \( \sin x + x \cos x = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ \sin x + x \cos x = 0 \]
Rearranging gives:
\[ \sin x = -x \cos x \]

### Step 2: Dividing by Cosine
We can divide both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ \tan x = -x \]

### Step 3: Analyzing the Functions
We need to analyze the functions \( y = \tan x \) and \( y = -x \). The function \( \tan x \) is periodic and has vertical asymptotes at \( x = \frac{\pi}{2} + n\pi \) for any integer \( n \). The function \( y = -x \) is a straight line with a slope of -1.

### Step 4: Finding Intersections
We need to find the points where these two functions intersect.

1. At \( x = 0 \):
\[ \tan(0) = 0 \quad \text{and} \quad -0 = 0 \]
So, \( x = 0 \) is one solution.

2. For \( x > 0 \):
- As \( x \) approaches \( \frac{\pi}{2} \), \( \tan x \) approaches \( +\infty \).
- The line \( y = -x \) will intersect the \( y \)-axis at \( 0 \) and decrease indefinitely.

3. For \( x < 0 \):
- As \( x \) approaches \( -\frac{\pi}{2} \), \( \tan x \) approaches \( -\infty \).
- The line \( y = -x \) will intersect the \( y \)-axis at \( 0 \) and increase indefinitely.

### Step 5: Applying the Intermediate Value Theorem
Since \( \tan x \) is continuous and crosses the line \( y = -x \) at least once in the intervals \( (-\pi, 0) \) and \( (0, \pi) \), we can conclude that there are roots in these intervals.

### Conclusion
Thus, the equation \( \sin x + x \cos x = 0 \) has at least one root in the intervals \( (-\pi, 0) \) and \( (0, \pi) \).

### Final Answer
The equation has at least one root in the interval \( (0, \pi) \).

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