Figure 2 shows a sketch of the curve C with parametric equations $x = 4 ext{sin} \left( t + \frac{\pi}{6} \right)$, $y = 3\text{cos} 2t$, $0 \leq t < 2\pi$ - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 8

To find where $\frac{dy}{dx} = 0$, we need to set the numerator of the previous expression to zero, since for a fraction to be zero, only the numerator must equal zero.

1. Set the numerator equal to zero: $-3\text{sin} 2t = 0$ This implies that: $\text{sin} 2t = 0$

2. Solve for $2t$: $2t = n\pi, \quad n \in \mathbb{Z}$ Thus, we have: $t = \frac{n\pi}{2}$

3. Since $0 \leq t < 2\pi$, valid integer values for $n$ are 0, 1, 2, 3, and 4:

   • For $n=0$: $t = 0$
   • For $n=1$: $t = \frac{\pi}{2}$
   • For $n=2$: $t = \pi$
   • For $n=3$: $t = \frac{3\pi}{2}$
   • For $n=4$: $t = 2\pi$

4. Now we find the corresponding coordinates:

   • For $t=0$: $x = 4\text{sin} \left( 0 + \frac{\pi}{6} \right) = 4\cdot\frac{1}{2} = 2$ $y = 3\text{cos} 0 = 3$ Coordinate: $(2, 3)$
   • For $t=\frac{\pi}{2}$: $x = 4\text{sin} \left( \frac{\pi}{2} + \frac{\pi}{6} \right) = 4\cdot\frac{\sqrt{3}}{2} = 2\sqrt{3}$ $y = 3\text{cos} \pi = -3$ Coordinate: $(2\sqrt{3}, -3)$
   • For $t=\pi$: $x = 4\text{sin} \left( \pi + \frac{\pi}{6} \right) = 4\cdot\left(-\frac{1}{2}\right) = -2$ $y = 3\text{cos} 2\pi = 3$ Coordinate: $(-2, 3)$
   • For $t=\frac{3\pi}{2}$: $x = 4\text{sin} \left( \frac{3\pi}{2} + \frac{\pi}{6} \right) = 4\cdot\left(-\frac{\sqrt{3}}{2}\right) = -2\sqrt{3}$ $y = 3\text{cos} 3\pi = -3$ Coordinate: $(-2\sqrt{3}, -3)$
   • For $t=2\pi$: $x = 4\text{sin} \left( 2\pi + \frac{\pi}{6} \right) = 4\cdot\frac{1}{2} = 2$ $y = 3\text{cos} 4\pi = 3$ Coordinate: $(2, 3)$

5. Summary of coordinates where $\frac{dy}{dx} = 0$:

   • $(2, 3)$
   • $(2\sqrt{3}, -3)$
   • $(-2, 3)$
   • $(-2\sqrt{3}, -3)$