Figure 1 shows an oscilloscope screen - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 6

Given the equation:

$y = 2 \sin(x + \frac{\pi}{6})$

To find when $y = 1$:

1. Set the equation equal to 1: $2 \sin(x + \frac{\pi}{6}) = 1$

2. Solving this gives: $\sin(x + \frac{\pi}{6}) = \frac{1}{2}$

3. The general solution for $\sin \theta = \frac{1}{2}$ is: $\theta = \frac{\pi}{6} + 2k\pi \\quad \text{or} \\quad \theta = \frac{5\pi}{6} + 2k\pi$ for integer k.

4. Substituting back for $x$:

   • For $\frac{\pi}{6}$: $x + \frac{\pi}{6} = \frac{\pi}{6} \\Rightarrow x = 0$
   • For $\frac{5\pi}{6}$: $x + \frac{\pi}{6} = \frac{5\pi}{6} \\Rightarrow x = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$
   • For $\frac{\pi}{6} + 2\pi$: $x + \frac{\pi}{6} = \frac{13\pi}{6} \\Rightarrow x = \frac{13\pi}{6} - \frac{\pi}{6} = 2\pi$

The valid values of $x$ in the range $0 ≤ x < 2π$ are: $x = 0, \frac{2\pi}{3}, 2\pi$