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

Options

A. ` alpha =0`
B. `alpha =0,-pi,-(pi)/(3)`
C. `alpha =(pi)/(6),(5pi)/(6)`
D. None of the above

Answer

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

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

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

Now, we apply the limits from \( \alpha \) to \( 2\alpha \):
\[ \int_{\alpha}^{2\alpha} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{\alpha}^{2\alpha} = \frac{1}{2} \sin(4\alpha) - \frac{1}{2} \sin(2\alpha) \]

### Step 2: Substitute the Integral Back into the Equation
Now we substitute the evaluated integral back into the original equation:
\[ \sin \alpha + \left( \frac{1}{2} \sin(4\alpha) - \frac{1}{2} \sin(2\alpha) \right) = 0 \]
This simplifies to:
\[ \sin \alpha + \frac{1}{2} \sin(4\alpha) - \frac{1}{2} \sin(2\alpha) = 0 \]

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

### Step 4: Analyze the Function
We will analyze the function:
\[ f(\alpha) = \sin \alpha + \frac{1}{2} \sin(4\alpha) - \frac{1}{2} \sin(2\alpha) \]
We need to find the number of roots of \( f(\alpha) = 0 \) in the interval \([- \pi, 0]\).

### Step 5: Check Specific Values
We can check specific values of \( \alpha \) in the interval to find the roots:
1. **At \( \alpha = -\pi \)**:
\[ f(-\pi) = \sin(-\pi) + \frac{1}{2} \sin(-4\pi) - \frac{1}{2} \sin(-2\pi) = 0 + 0 - 0 = 0 \]
2. **At \( \alpha = -\frac{\pi}{2} \)**:
\[ f\left(-\frac{\pi}{2}\right) = \sin\left(-\frac{\pi}{2}\right) + \frac{1}{2} \sin(-2\pi) - \frac{1}{2} \sin(-\pi) = -1 + 0 + 0 = -1 \]
3. **At \( \alpha = 0 \)**:
\[ f(0) = \sin(0) + \frac{1}{2} \sin(0) - \frac{1}{2} \sin(0) = 0 \]

### Step 6: Intermediate Value Theorem
Since \( f(-\pi) = 0 \) and \( f(-\frac{\pi}{2}) = -1 \), and \( f(0) = 0 \), we can conclude that there are at least two values of \( \alpha \) in the interval \([- \pi, 0]\) where \( f(\alpha) = 0\).

### Step 7: Conclusion
Thus, the number of values of \( \alpha \) in the interval \([- \pi, 0]\) satisfying the equation is **2**.

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