| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Factorising product of trig factors |
| Difficulty | Moderate -0.3 This is a straightforward C2 trigonometric equation using the null factor law. Part (a) requires simple factorization to get tan θ = -1 or tan²θ = 3, and part (b) involves standard angle-finding in a given interval. While it requires multiple steps and converting sin²θ - 3cos²θ = 0 using the identity tan²θ = sin²θ/cos²θ, these are routine C2 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = -1\) | B1 | |
| \(\frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta\), so \(\tan^2\theta = 3\) | M1, A1 | \(\frac{\sin\theta}{\cos\theta} = \tan\theta\) used on \(\sin^2\theta - 3\cos^2\theta\) |
| \(\tan\theta = \pm\sqrt{3}\) | A1 (4 marks) | Both values required |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = -1\), \(\tan\theta = \sqrt{3}\), \(\tan\theta = -\sqrt{3}\) | M1 | Uses part (a) to solve \(\tan\theta = k\) |
| \(\theta = 135°\), \(\theta = 60°\), \(\theta = 120°\) | A2,1,0 (3 marks) | If not A2 for all three correct, award A1 for two values correct; ignore answers outside \(0 \leq \theta \leq 180\) |
## Question 7:
### Part (a):
| $\tan\theta = -1$ | B1 | |
| $\frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$, so $\tan^2\theta = 3$ | M1, A1 | $\frac{\sin\theta}{\cos\theta} = \tan\theta$ used on $\sin^2\theta - 3\cos^2\theta$ |
| $\tan\theta = \pm\sqrt{3}$ | A1 (4 marks) | Both values required |
### Part (b):
| $\tan\theta = -1$, $\tan\theta = \sqrt{3}$, $\tan\theta = -\sqrt{3}$ | M1 | Uses part (a) to solve $\tan\theta = k$ |
| $\theta = 135°$, $\theta = 60°$, $\theta = 120°$ | A2,1,0 (3 marks) | If not A2 for all three correct, award A1 for two values correct; ignore answers outside $0 \leq \theta \leq 180$ |
---7 It is given that $( \tan \theta + 1 ) \left( \sin ^ { 2 } \theta - 3 \cos ^ { 2 } \theta \right) = 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the possible values of $\tan \theta$.
\item Hence solve the equation $( \tan \theta + 1 ) \left( \sin ^ { 2 } \theta - 3 \cos ^ { 2 } \theta \right) = 0$, giving all solutions for $\theta$, in degrees, in the interval $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q7 [7]}}