AQA C2 2008 June — Question 9 8 marks

Exam BoardAQA
ModuleC2 (Core Mathematics 2)
Year2008
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrigonometric equations in context
TypeConvert sin/cos ratio to tan
DifficultyModerate -0.3 Part (a) requires understanding that sin(2x) = sin(48°) leads to 2x = 48° or 2x = 180° - 48°, then finding all solutions in the doubled range—a standard technique. Part (b) involves rearranging to tan θ = 3/2 and finding solutions in two quadrants. Both are routine C2-level trigonometric equation problems requiring familiar methods without novel insight, making this slightly easier than average.
Spec1.05o Trigonometric equations: solve in given intervals

9
  1. Solve the equation \(\sin 2 x = \sin 48 ^ { \circ }\), giving the values of \(x\) in the interval \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).
  2. Solve the equation \(2 \sin \theta - 3 \cos \theta = 0\) in the interval \(0 ^ { \circ } \leqslant \theta < 360 ^ { \circ }\), giving your answers to the nearest \(0.1 ^ { \circ }\).

AnswerMarks Guidance
(a) \(2x = 48\); \(2x = 180 - 48\); \(2x = 360 + 48\) and \(2x = 360 + 180 - 48\); \(x = 24°, 66°, 204°, 246°\)B1, M1, M1, M1, A1 (4 marks) PI by \(x = 24°\); Accept equivalents for \(x\); Accept equivalents for \(x\); CAO; need all four, no extras in given interval
(b) \(\frac{\sin\theta}{\cos\theta} = \tan\theta\); \(2\sin\theta - 3\cos\theta = 0 \Rightarrow \tan\theta = 1.5\); \(\theta = 56.3°\); \(\theta = 56.3° + 180° = 236.3°\)M1, A1, A1, A1F (4 marks) Stated or used; Condone > 1dp; Ft on \(c's\) PV+180° dep only on the M1 provided no 'extra' solutions in the given interval.
Total for Q9: 8 marks
TOTAL: 75 marks
**(a)** $2x = 48$; $2x = 180 - 48$; $2x = 360 + 48$ and $2x = 360 + 180 - 48$; $x = 24°, 66°, 204°, 246°$ | B1, M1, M1, M1, A1 (4 marks) | PI by $x = 24°$; Accept equivalents for $x$; Accept equivalents for $x$; CAO; need all four, no extras in given interval

**(b)** $\frac{\sin\theta}{\cos\theta} = \tan\theta$; $2\sin\theta - 3\cos\theta = 0 \Rightarrow \tan\theta = 1.5$; $\theta = 56.3°$; $\theta = 56.3° + 180° = 236.3°$ | M1, A1, A1, A1F (4 marks) | Stated or used; Condone > 1dp; Ft on $c's$ PV+180° dep only on the M1 provided no 'extra' solutions in the given interval.

**Total for Q9: 8 marks**

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## TOTAL: 75 marks
9
\begin{enumerate}[label=(\alph*)]
\item Solve the equation $\sin 2 x = \sin 48 ^ { \circ }$, giving the values of $x$ in the interval $0 ^ { \circ } \leqslant x < 360 ^ { \circ }$.
\item Solve the equation $2 \sin \theta - 3 \cos \theta = 0$ in the interval $0 ^ { \circ } \leqslant \theta < 360 ^ { \circ }$, giving your answers to the nearest $0.1 ^ { \circ }$.
\end{enumerate}

\hfill \mbox{\textit{AQA C2 2008 Q9 [8]}}