| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | General solution — then find specific solutions |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths trig equation requiring standard technique: isolate sin, find principal value, apply general solution formula for the transformed argument, then substitute to find a specific solution. While it's FP1, the method is routine and mechanical with no conceptual challenges beyond A-level core content. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\sin(3x + 45°) = \frac{1}{2}\) | M1 | Dividing both sides by 2 |
| \(3x + 45° = 30°\) or \(150°\) | B1 | At least one correct principal value |
| \(3x + 45° = 180n° + (-1)^n \cdot 30°\) | M1 | Correct general solution form using either value |
| \(3x = 180n° + (-1)^n \cdot 30° - 45°\) | A1 | Correct unsimplified general solution |
| \(x = 60n° + (-1)^n \cdot 10° - 15°\) | A1 | Correct general solution in \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Substituting values of \(n\) to find solution closest to \(200°\), giving \(x = 195°\) | B1 | Follow through from general solution in (a) |
## Question 4:
**Part (a):** Find the general solution of $2\sin(3x + 45°) = 1$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\sin(3x + 45°) = \frac{1}{2}$ | M1 | Dividing both sides by 2 |
| $3x + 45° = 30°$ or $150°$ | B1 | At least one correct principal value |
| $3x + 45° = 180n° + (-1)^n \cdot 30°$ | M1 | Correct general solution form using either value |
| $3x = 180n° + (-1)^n \cdot 30° - 45°$ | A1 | Correct unsimplified general solution |
| $x = 60n° + (-1)^n \cdot 10° - 15°$ | A1 | Correct general solution in $x$ |
**Part (b):** Solution closest to 200°
| Working/Answer | Mark | Guidance |
|---|---|---|
| Substituting values of $n$ to find solution closest to $200°$, giving $x = 195°$ | B1 | Follow through from general solution in (a) |
---4
\begin{enumerate}[label=(\alph*)]
\item Find the general solution, in degrees, of the equation
$$2 \sin \left( 3 x + 45 ^ { \circ } \right) = 1$$
\item Use your general solution to find the solution of $2 \sin \left( 3 x + 45 ^ { \circ } \right) = 1$ that is closest to $200 ^ { \circ }$.\\[0pt]
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2015 Q4 [6]}}