`sinx=sinx^(@)` if `x=`

Options

A. `(180pi)/(180+pi)`
B. `(180pi)/(180-pi)`
C. `(360pi)/(180-pi)`
D. `(360pi)/(180+pi)`

Answer

To solve the equation \( \sin x = \sin x^\circ \), we will first convert \( x^\circ \) into radians and then apply the properties of the sine function.

### Step-by-step Solution:

1. **Convert Degrees to Radians**:
We know that \( 180^\circ \) is equivalent to \( \pi \) radians. Therefore, to convert \( x^\circ \) to radians, we use the conversion factor:
\[ x^\circ = \frac{\pi x}{180} \text{ radians} \]

2. **Set Up the Equation**:
Now we can rewrite the equation:
\[ \sin x = \sin\left(\frac{\pi x}{180}\right) \]

3. **Use the Sine Function Identity**:
The sine function has the property that \( \sin A = \sin B \) implies:
\[ A = n\pi + (-1)^n B \quad \text{for } n \in \mathbb{Z} \]
Applying this to our equation gives:
\[ x = n\pi + (-1)^n \left(\frac{\pi x}{180}\right) \]

4. **Solve for \( x \)**:
Let's consider two cases for \( n \).

- **Case 1**: Let \( n = 1 \)
\[ x = \pi + \left(-1\right)\left(\frac{\pi x}{180}\right) \]
Rearranging gives:
\[ x + \frac{\pi x}{180} = \pi \]
Factoring out \( x \):
\[ x\left(1 + \frac{\pi}{180}\right) = \pi \]
Thus,
\[ x = \frac{\pi}{1 + \frac{\pi}{180}} = \frac{180\pi}{180 + \pi} \]

- **Case 2**: Let \( n = 2 \)
\[ x = 2\pi + \left(-1\right)\left(\frac{\pi x}{180}\right) \]
Rearranging gives:
\[ x + \frac{\pi x}{180} = 2\pi \]
Factoring out \( x \):
\[ x\left(1 + \frac{\pi}{180}\right) = 2\pi \]
Thus,
\[ x = \frac{2\pi}{1 + \frac{\pi}{180}} = \frac{360\pi}{360 + \pi} \]

5. **Final Solutions**:
The solutions for \( x \) are:
\[ x = \frac{180\pi}{180 + \pi} \quad \text{and} \quad x = \frac{360\pi}{360 + \pi} \]