At any two points of the curve represented parametrically by `x=a (2 cos t- cos 2t);y = a (2 sin t - sin 2t)` the tangents are parallel to the axis of x corresponding to the values of the parameter t differing from each other by :

Options

A. `2pi//3`
B. `3pi//4`
C. `pi//2`
D. `pi//3`

Answer

To solve the problem, we need to determine the values of the parameter \( t \) for which the tangents to the curve defined by the parametric equations \( x = a(2 \cos t - \cos 2t) \) and \( y = a(2 \sin t - \sin 2t) \) are parallel to the x-axis. This occurs when the slope of the tangent (given by \( \frac{dy}{dx} \)) is equal to zero.

### Step-by-Step Solution

1. **Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)**:
- Differentiate \( x \) with respect to \( t \):
\[ \frac{dx}{dt} = a \left( -2 \sin t + 2 \sin 2t \right) = a \left( -2 \sin t + 2 \cdot 2 \sin t \cos t \right) = a \left( -2 \sin t + 4 \sin t \cos t \right) \]
- Differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = a \left( 2 \cos t - 2 \cos 2t \right) = a \left( 2 \cos t - 2 \cdot (1 - 2 \sin^2 t) \right) = a \left( 2 \cos t - 2 + 4 \sin^2 t \right) \]

2. **Find the slope \( \frac{dy}{dx} \)**:
- The slope of the tangent line is given by:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]
- Set this equal to zero for the tangents to be parallel to the x-axis:
\[ \frac{dy}{dx} = 0 \implies \frac{dy}{dt} = 0 \]

3. **Set \( \frac{dy}{dt} = 0 \)**:
- From our previous expression for \( \frac{dy}{dt} \):
\[ 2 \cos t - 2 + 4 \sin^2 t = 0 \]
- Rearranging gives:
\[ 2 \cos t + 4 \sin^2 t = 2 \]
- Using the identity \( \sin^2 t = 1 - \cos^2 t \):
\[ 2 \cos t + 4(1 - \cos^2 t) = 2 \]
- Simplifying:
\[ 2 \cos t + 4 - 4 \cos^2 t = 2 \implies -4 \cos^2 t + 2 \cos t + 2 = 0 \]
- Dividing by -2:
\[ 2 \cos^2 t - \cos t - 1 = 0 \]

4. **Solve the quadratic equation**:
- Using the quadratic formula \( \cos t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \cos t = \frac{1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \]
- This gives:
\[ \cos t = 1 \quad \text{or} \quad \cos t = -\frac{1}{2} \]

5. **Find corresponding values of \( t \)**:
- For \( \cos t = 1 \):
\[ t = 2n\pi \quad (n \in \mathbb{Z}) \]
- For \( \cos t = -\frac{1}{2} \):
\[ t = \frac{2\pi}{3} + 2n\pi \quad \text{or} \quad t = \frac{4\pi}{3} + 2n\pi \]

6. **Calculate the difference in \( t \)**:
- The difference between \( t = 2n\pi \) and \( t = \frac{2\pi}{3} \) or \( t = \frac{4\pi}{3} \) is:
\[ \Delta t = \left| 2n\pi - \frac{2\pi}{3} \right| = \frac{2\pi}{3} \quad \text{or} \quad \Delta t = \left| 2n\pi - \frac{4\pi}{3} \right| = \frac{4\pi}{3} \]

### Final Result
The values of the parameter \( t \) differing from each other by \( \frac{2\pi}{3} \) are the ones we are interested in. Thus, the answer is:

\[ \boxed{\frac{2\pi}{3}} \]