AQA FP1 2011 January — Question 4 6 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeGeneral solution — find all solutions
DifficultyStandard +0.3 This is a straightforward Further Maths trigonometric equation requiring knowledge of general solutions and the unit circle. Students must find where sin equals -1/2, apply the general solution formula for sine, then solve for x. While it's FP1 content, it's a standard textbook exercise with clear methodology and no novel problem-solving required, making it slightly easier than average overall.
Spec1.05o Trigonometric equations: solve in given intervals

4 Find the general solution of the equation $$\sin \left( 4 x - \frac { 2 \pi } { 3 } \right) = - \frac { 1 } { 2 }$$ giving your answer in terms of \(\pi\).
(6 marks)

AnswerMarks Guidance
\(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\)B1
\(\sin\left(-\frac{5\pi}{6}\right) = -\frac{1}{2}\)B1F
Use of \(2\pi\)M1
Going from \(4x - \frac{2\pi}{3}\) to \(x\)m1
GS \(x = \frac{\pi}{8} + \frac{1}{2}n\pi\) or \(x = -\frac{5\pi}{24} + \frac{1}{2}n\pi\)A1A1 6 marks
Total: 6 marks
$\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$ | B1 | | OE; dec/deg penalised at 6th mark

$\sin\left(-\frac{5\pi}{6}\right) = -\frac{1}{2}$ | B1F | | OE; ft wrong first value

Use of $2\pi$ | M1 | | (or $n\pi$) at any stage

Going from $4x - \frac{2\pi}{3}$ to $x$ | m1 | | including division of all terms by 4

GS $x = \frac{\pi}{8} + \frac{1}{2}n\pi$ or $x = -\frac{5\pi}{24} + \frac{1}{2}n\pi$ | A1A1 | 6 marks | OE

**Total: 6 marks**

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4 Find the general solution of the equation

$$\sin \left( 4 x - \frac { 2 \pi } { 3 } \right) = - \frac { 1 } { 2 }$$

giving your answer in terms of $\pi$.\\
(6 marks)

\hfill \mbox{\textit{AQA FP1 2011 Q4 [6]}}