AQA FP1 2009 January — Question 3 5 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeGeneral solution — find all solutions
DifficultyModerate -0.5 This is a straightforward trigonometric equation requiring knowledge that tan(π/2 - θ) = cot(θ) = 1/tan(θ), recognition that tan(π/3) = √3, and application of the general solution formula for tan. While it's Further Maths content, it's a routine single-step problem with standard technique application, making it slightly easier than average.
Spec1.05o Trigonometric equations: solve in given intervals
3 Find the general solution of the equation $$\tan \left( \frac { \pi } { 2 } - 3 x \right) = \sqrt { 3 }$$
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\tan\dfrac{\pi}{3} = \sqrt{3}\)B1 Decimals/degrees penalised at 5th mark
Introduction of \(n\pi\)M1 (or \(2n\pi\)) at any stage
Going from \(\dfrac{\pi}{2}-3x\) to \(x\)m1 Including dividing all terms by 3
\(x = \dfrac{\pi}{18}+\dfrac{1}{3}n\pi\)A2,1F Allow \(+,-\) or \(\pm\); A1 with dec/deg; ft wrong first solution
Total5 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\dfrac{\pi}{3} = \sqrt{3}$ | B1 | Decimals/degrees penalised at 5th mark |
| Introduction of $n\pi$ | M1 | (or $2n\pi$) at any stage |
| Going from $\dfrac{\pi}{2}-3x$ to $x$ | m1 | Including dividing all terms by 3 |
| $x = \dfrac{\pi}{18}+\dfrac{1}{3}n\pi$ | A2,1F | Allow $+,-$ or $\pm$; A1 with dec/deg; ft wrong first solution |
| **Total** | **5** | |

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3 Find the general solution of the equation

$$\tan \left( \frac { \pi } { 2 } - 3 x \right) = \sqrt { 3 }$$

\hfill \mbox{\textit{AQA FP1 2009 Q3 [5]}}
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Q1 5 Q2 5 Q3 5 Q4 7 Q5 12 Q6 10 Q7 10 Q8 7 Q9 14