| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| 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 question requiring standard technique for general solutions of trigonometric equations with a linear transformation of the variable. While it involves radians and π, the method is routine: solve for the transformed angle, then back-substitute. The 'hence' part in (b) is trivial once (a) is complete. Slightly easier than average due to its mechanical nature. |
| Spec | 1.05g Exact trigonometric values: for standard angles1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\) stated or used; Appropriate use of \(\pm\); Introduction of \(2n\pi\); Subtraction of \(\frac{\pi}{3}\) and multiplication by 2; \(x = -\frac{2\pi}{3} + \frac{\pi}{2} + 4n\pi\) | B1, B1, M1, m1, A1 | 5 marks |
| (b) \(n = 1\) gives min pos \(x = \frac{17\pi}{6}\) | M1A1 | 2 marks |
**(a)** $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ stated or used; Appropriate use of $\pm$; Introduction of $2n\pi$; Subtraction of $\frac{\pi}{3}$ and multiplication by 2; $x = -\frac{2\pi}{3} + \frac{\pi}{2} + 4n\pi$ | B1, B1, M1, m1, A1 | 5 marks | Degrees or decimals penalised in 5th mark only; OE; OE; All terms multiplied by 2; OE
**(b)** $n = 1$ gives min pos $x = \frac{17\pi}{6}$ | M1A1 | 2 marks | NMS 1/2 provided (a) correct
**Total: 7 marks**5
\begin{enumerate}[label=(\alph*)]
\item Find, in radians, the general solution of the equation
$$\cos \left( \frac { x } { 2 } + \frac { \pi } { 3 } \right) = \frac { 1 } { \sqrt { 2 } }$$
giving your answer in terms of $\pi$.
\item Hence find the smallest positive value of $x$ which satisfies this equation.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2008 Q5 [7]}}