The value of `theta` in the interval `[- pi, 0]` satisfying `sin theta + int_(theta)^(2theta) cos 2x dx = 0` is

Options

A. `-pi//2`
B. `-pi`
C. `-pi//3`
D. `0`

Answer

To solve the equation \( \sin \theta + \int_{\theta}^{2\theta} \cos 2x \, dx = 0 \) for \( \theta \) in the interval \([- \pi, 0]\), we will follow these steps:

### Step 1: Evaluate the integral
We need to evaluate the integral \( \int_{\theta}^{2\theta} \cos 2x \, dx \).

Using the integral formula:
\[ \int \cos kx \, dx = \frac{1}{k} \sin kx + C \]
we have:
\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x + C \]

Thus, we can evaluate the definite integral:
\[ \int_{\theta}^{2\theta} \cos 2x \, dx = \left[ \frac{1}{2} \sin 2x \right]_{\theta}^{2\theta} = \frac{1}{2} \sin(4\theta) - \frac{1}{2} \sin(2\theta) \]

### Step 2: Substitute the integral back into the equation
Now substitute this result back into the original equation:
\[ \sin \theta + \left( \frac{1}{2} \sin(4\theta) - \frac{1}{2} \sin(2\theta) \right) = 0 \]
This simplifies to:
\[ \sin \theta + \frac{1}{2} \sin(4\theta) - \frac{1}{2} \sin(2\theta) = 0 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ \sin \theta + \frac{1}{2} \sin(4\theta) = \frac{1}{2} \sin(2\theta) \]

### Step 4: Testing possible values of \( \theta \)
Now we will test the possible values of \( \theta \) in the interval \([- \pi, 0]\) to find which satisfies the equation.

1. **Testing \( \theta = -\pi \)**:
\[ \sin(-\pi) + \frac{1}{2} \sin(-4\pi) - \frac{1}{2} \sin(-2\pi) = 0 + 0 - 0 = 0 \]
This satisfies the equation.

2. **Testing \( \theta = -\frac{\pi}{3} \)**:
\[ \sin(-\frac{\pi}{3}) + \frac{1}{2} \sin(-\frac{4\pi}{3}) - \frac{1}{2} \sin(-\frac{2\pi}{3}) \]
\[ = -\frac{\sqrt{3}}{2} + \frac{1}{2} \left(-\frac{\sqrt{3}}{2}\right) - \frac{1}{2} \left(\frac{\sqrt{3}}{2}\right) \]
\[ = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\sqrt{3} \neq 0 \]
This does not satisfy the equation.

3. **Testing \( \theta = -\frac{\pi}{2} \)**:
\[ \sin(-\frac{\pi}{2}) + \frac{1}{2} \sin(-2\pi) - \frac{1}{2} \sin(-\pi) \]
\[ = -1 + 0 - 0 = -1 \neq 0 \]
This does not satisfy the equation.

4. **Testing \( \theta = 0 \)**:
\[ \sin(0) + \frac{1}{2} \sin(0) - \frac{1}{2} \sin(0) = 0 + 0 - 0 = 0 \]
This satisfies the equation.

### Conclusion
The values of \( \theta \) that satisfy the equation in the interval \([- \pi, 0]\) are \( \theta = -\pi \) and \( \theta = 0 \).