| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | General solution — find all solutions |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths question requiring knowledge of general solutions for tan equations and the principal value of tan(√3). Part (a) is direct substitution, part (b) adds a simple phase shift. While it's FP1 content, the technique is mechanical once the pattern is learned, making it slightly easier than average overall. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\tan \frac{\pi}{3} = \sqrt{3}\); GS is \(\frac{1}{3}(\frac{\pi}{3} + n\pi)\) | M1, m1A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| One solution is \(0\); GS is \(\frac{1}{3}n\pi\) | M1A1, m1A1F | 4 marks |
**Part (a)**
| Use of $\tan \frac{\pi}{3} = \sqrt{3}$; GS is $\frac{1}{3}(\frac{\pi}{3} + n\pi)$ | M1, m1A1 | 3 marks | Degrees used - 1 mark penalty; m1A0 for $\frac{\pi}{6} + n\pi$ or for correct answer plus extra solutions |
**Part (b)**
| One solution is $0$; GS is $\frac{1}{3}n\pi$ | M1A1, m1A1F | 4 marks | OE; m1A0 for $n\pi$ or for correct answer plus extra solns; ft wrong first solution |
**Total: 7 marks**
---5 Find the general solutions of the following equations, giving your answers in terms of $\pi$ :
\begin{enumerate}[label=(\alph*)]
\item $\quad \tan 3 x = \sqrt { 3 }$;
\item $\quad \tan \left( 3 x - \frac { \pi } { 3 } \right) = - \sqrt { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2005 Q5 [7]}}