For the curve `y=3sinthetacostheta, x=e^(theta)sintheta, 0lt=thetalt=pi,` the tangent is parallel to x-axis when `theta` is :

Options

A. 0
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `(pi)/(6)`

Answer

To solve the problem, we need to determine the value of \(\theta\) for which the tangent to the curve is parallel to the x-axis. This occurs when the derivative \(\frac{dy}{dx} = 0\).

### Step-by-Step Solution:

1. **Identify the given functions:**
\[ y = 3 \sin \theta \cos \theta \]
\[ x = e^{\theta} \sin \theta \]

2. **Find \(\frac{dy}{d\theta}\):**
We will use the product rule to differentiate \(y\):
\[ \frac{dy}{d\theta} = 3 \left( \sin \theta \frac{d}{d\theta}(\cos \theta) + \cos \theta \frac{d}{d\theta}(\sin \theta) \right) \]
\[ = 3 \left( \sin \theta (-\sin \theta) + \cos \theta (\cos \theta) \right) \]
\[ = 3 \left( \cos^2 \theta - \sin^2 \theta \right) \]

3. **Find \(\frac{dx}{d\theta}\):**
Again, we will use the product rule to differentiate \(x\):
\[ \frac{dx}{d\theta} = e^{\theta} \frac{d}{d\theta}(\sin \theta) + \sin \theta \frac{d}{d\theta}(e^{\theta}) \]
\[ = e^{\theta} \cos \theta + \sin \theta e^{\theta} \]
\[ = e^{\theta} (\cos \theta + \sin \theta) \]

4. **Find \(\frac{dy}{dx}\):**
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3(\cos^2 \theta - \sin^2 \theta)}{e^{\theta}(\cos \theta + \sin \theta)} \]

5. **Set \(\frac{dy}{dx} = 0\):**
For the tangent to be parallel to the x-axis:
\[ \frac{dy}{dx} = 0 \implies 3(\cos^2 \theta - \sin^2 \theta) = 0 \]
\[ \cos^2 \theta - \sin^2 \theta = 0 \]
\[ \cos^2 \theta = \sin^2 \theta \]

6. **Solve for \(\theta\):**
This implies:
\[ \cos \theta = \sin \theta \]
This equality holds true when:
\[ \theta = \frac{\pi}{4} \]

7. **Check the interval:**
Since \(0 < \theta \leq \pi\), \(\theta = \frac{\pi}{4}\) is within the specified range.

### Final Answer:
The value of \(\theta\) for which the tangent to the curve is parallel to the x-axis is:
\[ \theta = \frac{\pi}{4} \]