The equation of the tangent to the curve `y=2 sin x +sin 2x` at `x=pi/3` is equal to

Options

A. `2y=3sqrt3`
B. `y=3sqrt3`
C. `2y+3sqrt3=0`
D. `y+3sqrt3=0`

Answer

To find the equation of the tangent to the curve \( y = 2 \sin x + \sin 2x \) at \( x = \frac{\pi}{3} \), we will follow these steps:

### Step 1: Find the y-coordinate at \( x = \frac{\pi}{3} \)
We need to substitute \( x = \frac{\pi}{3} \) into the equation of the curve to find the corresponding y-coordinate.

\[ y = 2 \sin\left(\frac{\pi}{3}\right) + \sin\left(2 \cdot \frac{\pi}{3}\right) \]

Calculating \( \sin\left(\frac{\pi}{3}\right) \) and \( \sin\left(\frac{2\pi}{3}\right) \):

\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
\[ \sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Substituting these values back into the equation:

\[ y = 2 \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

So, the point on the curve at \( x = \frac{\pi}{3} \) is \( \left(\frac{\pi}{3}, \frac{3\sqrt{3}}{2}\right) \).

### Step 2: Find the derivative \( \frac{dy}{dx} \) to get the slope of the tangent
Now, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = 2 \cos x + 2 \cos 2x \]

### Step 3: Evaluate the derivative at \( x = \frac{\pi}{3} \)
Substituting \( x = \frac{\pi}{3} \) into the derivative:

\[ \frac{dy}{dx} = 2 \cos\left(\frac{\pi}{3}\right) + 2 \cos\left(2 \cdot \frac{\pi}{3}\right) \]

Calculating \( \cos\left(\frac{\pi}{3}\right) \) and \( \cos\left(\frac{2\pi}{3}\right) \):

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]

Substituting these values back into the derivative:

\[ \frac{dy}{dx} = 2 \cdot \frac{1}{2} + 2 \cdot \left(-\frac{1}{2}\right) = 1 - 1 = 0 \]

### Step 4: Write the equation of the tangent line
The slope \( m \) of the tangent line is \( 0 \). The equation of the tangent line at the point \( \left(\frac{\pi}{3}, \frac{3\sqrt{3}}{2}\right) \) is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting \( y_1 = \frac{3\sqrt{3}}{2} \), \( m = 0 \), and \( x_1 = \frac{\pi}{3} \):

\[ y - \frac{3\sqrt{3}}{2} = 0 \implies y = \frac{3\sqrt{3}}{2} \]

### Step 5: Convert to the required form
The equation \( y = \frac{3\sqrt{3}}{2} \) can be rewritten as:

\[ 2y = 3\sqrt{3} \]

Thus, the equation of the tangent to the curve at \( x = \frac{\pi}{3} \) is:

\[ \boxed{2y = 3\sqrt{3}} \]