| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Multiple independent equations — includes show/prove component |
| Difficulty | Standard +0.3 This is a standard C2 trigonometric equation question requiring routine techniques: solving tan x directly, converting a trig equation to quadratic form by dividing by cos²θ (a bookwork manipulation), then solving the quadratic and finding angles. While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reference angle: \(\arctan(3) = 71.6°\) | M1 | Finding reference angle |
| \(x = 180° - 71.6° = 108.4°\) | A1 | First solution |
| \(x = 360° - 71.6° = 288.4°\) | A1 | Second solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(7\sin^2\theta + \sin\theta\cos\theta = 6\) | ||
| Divide through by \(\cos^2\theta\): \(7\tan^2\theta + \tan\theta = \frac{6}{\cos^2\theta}\) | M1 | Dividing by \(\cos^2\theta\) |
| Using \(\frac{1}{\cos^2\theta} = 1+\tan^2\theta\): \(7\tan^2\theta + \tan\theta = 6(1+\tan^2\theta)\) | M1 | Using \(\sec^2\theta = 1+\tan^2\theta\) |
| \(\tan^2\theta + \tan\theta - 6 = 0\) | A1 | Correct simplification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\tan\theta + 3)(\tan\theta - 2) = 0\) | M1 | Factorising |
| \(\tan\theta = -3\) or \(\tan\theta = 2\) | A1 | Both values |
| \(\tan\theta = -3\): \(\theta = 108.4°, 288.4°\) | A1 | Both solutions from part (a) |
| \(\tan\theta = 2\): \(\theta = 63.4°, 243.4°\) | A1 | Both solutions |
# Question 9:
## Part (a) – Solve $\tan x = -3$
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reference angle: $\arctan(3) = 71.6°$ | M1 | Finding reference angle |
| $x = 180° - 71.6° = 108.4°$ | A1 | First solution |
| $x = 360° - 71.6° = 288.4°$ | A1 | Second solution |
## Part (b)(i) – Show $\tan^2\theta + \tan\theta - 6 = 0$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $7\sin^2\theta + \sin\theta\cos\theta = 6$ | | |
| Divide through by $\cos^2\theta$: $7\tan^2\theta + \tan\theta = \frac{6}{\cos^2\theta}$ | M1 | Dividing by $\cos^2\theta$ |
| Using $\frac{1}{\cos^2\theta} = 1+\tan^2\theta$: $7\tan^2\theta + \tan\theta = 6(1+\tan^2\theta)$ | M1 | Using $\sec^2\theta = 1+\tan^2\theta$ |
| $\tan^2\theta + \tan\theta - 6 = 0$ | A1 | Correct simplification |
## Part (b)(ii) – Solve in $[0°, 360°]$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\tan\theta + 3)(\tan\theta - 2) = 0$ | M1 | Factorising |
| $\tan\theta = -3$ or $\tan\theta = 2$ | A1 | Both values |
| $\tan\theta = -3$: $\theta = 108.4°, 288.4°$ | A1 | Both solutions from part (a) |
| $\tan\theta = 2$: $\theta = 63.4°, 243.4°$ | A1 | Both solutions |9
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\tan x = - 3$ in the interval $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$, giving your answers to the nearest degree.
\item \begin{enumerate}[label=(\roman*)]
\item Given that
$$7 \sin ^ { 2 } \theta + \sin \theta \cos \theta = 6$$
show that
$$\tan ^ { 2 } \theta + \tan \theta - 6 = 0$$
\item Hence solve the equation $7 \sin ^ { 2 } \theta + \sin \theta \cos \theta = 6$ in the interval $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$, giving your answers to the nearest degree.\\
(4 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q9 [10]}}