Find the solutions of the equation sin(3x−15°)=12, for which 0≤x≤180° - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 4

Given coordinates for P, Q, and R are:

• $P(\frac{\pi}{10}, 0)$
• $Q(35°, 0)$
• $R(\frac{117°}{10}, 0)$

The equation for the curve is $y = \text{sin}(ax - b)$. Since these points cut the x-axis, it means that:

1. For point P:

   $$\text{sin}\left(a\frac{\pi}{10} - b\right) = 0$$

   Hence,

   $$a\frac{\pi}{10} - b = n\pi ext{ for integers } n$$

2. For point Q:

   $$\text{sin}(35°a - b) = 0$$

   Therefore,

   $$35°a - b = m\pi ext{ for integers } m$$

3. For point R:

   $$\text{sin}\left(a\frac{117°}{10} - b\right) = 0$$

   Thus,

   $$a\frac{117°}{10} - b = p\pi ext{ for integers } p$$

From these equations, we can solve for $a$ and $b$. Equating the first two equations allows us to isolate and equate the terms:

$$\frac{n\pi + b}{\frac{\pi}{10}} = a ext{ and } \frac{m\pi + b}{35°} = a \ \Rightarrow \frac{n\pi + b}{\frac{\pi}{10}} = \frac{m\pi + b}{35°}$$

Solving this can yield values for $a$ and $b$. This approach requires systematic calculation based on the relationships established.