| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: compound angle substitution |
| Difficulty | Standard +0.3 This is a standard C4 trigonometry question testing routine techniques: double angle formula manipulation, solving quadratic equations in sin x, and the R-formula method. All parts follow textbook procedures with no novel insight required. Part (a) is straightforward algebraic manipulation using cos 2x = 1 - 2sin²x, part (b) is factorizing a simple quadratic, and part (c) applies the standard R-formula technique. The multi-part structure and 11 total marks indicate moderate length, but each component is a well-practiced exam technique, making this slightly easier than the average A-level question. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos 2x = 1 - 2\sin^2 x\) | B1 | |
| \(3(1 - 2\sin^2 x) + 2\sin x + 1 = 0\) | M1 | |
| \(-6\sin^2 x + 2\sin x + 4 = 0\); \(3\sin^2 x - \sin x - 2 = 0\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \((3\sin x + 2)(\sin x - 1) = 0\) | M1 | |
| \(\sin x = -\frac{2}{3}\), \(\sin x = 1\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = \sqrt{13}\) | B1 | Accept 3.6 or better |
| \(\tan\alpha = \frac{2}{3}\), \(\alpha = 33.7°\) | M1 A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x - \alpha = \cos^{-1}\left(\frac{-1}{R}\right)\) | M1 | Candidate's \(R\). Or \(\cos(2x - \alpha) = \frac{-1}{R}\) |
| \(2x - \alpha = 106.1°, 253.9°\) | A1 | |
| \(x = 69.9°, 143.8°\) | A1 | 3 marks |
**5(a)(i)**
$\cos 2x = 1 - 2\sin^2 x$ | B1
$3(1 - 2\sin^2 x) + 2\sin x + 1 = 0$ | M1
$-6\sin^2 x + 2\sin x + 4 = 0$; $3\sin^2 x - \sin x - 2 = 0$ | A1 | 3 marks | AG
**5(a)(ii)**
$(3\sin x + 2)(\sin x - 1) = 0$ | M1
$\sin x = -\frac{2}{3}$, $\sin x = 1$ | A1 | 2 marks | Factorise correctly or use formula correctly; Both; condone $-0.67$ or $-0.66$ or better
**5(b)(i)**
$R = \sqrt{13}$ | B1 | Accept 3.6 or better
$\tan\alpha = \frac{2}{3}$, $\alpha = 33.7°$ | M1 A1 | 3 marks | OE; accept $\alpha = 33.69(0)$
**5(b)(ii)**
$2x - \alpha = \cos^{-1}\left(\frac{-1}{R}\right)$ | M1 | Candidate's $R$. Or $\cos(2x - \alpha) = \frac{-1}{R}$
$2x - \alpha = 106.1°, 253.9°$ | A1
$x = 69.9°, 143.8°$ | A1 | 3 marks | One correct answer; Both correct, no extras in range
---\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Show that the equation $3\cos 2x + 2\sin x + 1 = 0$ can be written in the form
$$3\sin^2 x - \sin x - 2 = 0$$ [3 marks]
\item Hence, given that $3\cos 2x + 2\sin x + 1 = 0$, find the possible values of $\sin x$. [2 marks]
\end{enumerate}
\item
\begin{enumerate}[label=(\roman*)]
\item Express $3\cos 2x + 2\sin 2x$ in the form $R\cos(2x - \alpha)$, where $R > 0$ and $0° < \alpha < 90°$, giving $\alpha$ to the nearest $0.1°$. [3 marks]
\item Hence solve the equation
$$3\cos 2x + 2\sin 2x + 1 = 0$$
for all solutions in the interval $0° < x < 180°$, giving $x$ to the nearest $0.1°$. [3 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2010 Q5 [11]}}