Edexcel PURE 2024 October — Question 1

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicReciprocal Trig & Identities
TypeSolve equation using Pythagorean identities
DifficultyStandard +0.3 This is a standard Further Maths Pure reciprocal trig equation requiring the identity sec²θ = 1 + tan²θ to convert to a quadratic in sec θ, then solving and finding angles. While it requires multiple steps (substitution, quadratic formula, inverse trig, finding all solutions in range), the technique is routine for FM students and follows a well-practiced procedure with no novel insight needed.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals
  1. In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Solve, for \(0 < \theta \leqslant 360 ^ { \circ }\), the equation $$3 \tan ^ { 2 } \theta + 7 \sec \theta - 3 = 0$$ giving your answers to one decimal place.
Question 1:
AnswerMarks
1( )
3tan2+7sec−3=03 sec 2−1 +7sec−3=0M1
3 s e c 2 +  s e c − 6 = 0  A1
( 3sec−)( sec+)=0sec=...cos=...dM1
=              A1, A1
(5)

Total 5

If a calculator is used the solution must be correct for their equation.
E.g. 6 c o s 2 −  c o s − 3 = 0  c o s = −
1
3
  
A1: One of = awrtawrt following M1, A1, dM1
Allow either answer in radians here so score for awrt 1.91 or awrt 4.37
A1: CSO. Both = awrtandawrt and no extras in the range.
If both solutions to the quadratic were given then both must be correct
………………………………………………………………………………………………………..
Methods using calculator technology and not showing all stages of working as required by question:
Example 1:
3tan2+7sec−3=0= 
No marks
Example 2:
3 t a n
3
2
s e c 2
7 s e c
s e c
3
6
0
0
3
(
s e
s
c
e
2
c
1
)
3
7 s e c 3 0    
    
+
+ 
− =
− =
−
= −
+
 = 
−
  
=
        
Scores M1, A1, dM0, A0, A0
Example 3:
3 t a n 2 + 7 s e c − 3 = 0  6 c o s 2 −  c o s − 3 = 0     via identities followed by  =              
Scores M1, A1, dM0, A0, A0
Methods using calculator technology and showing sufficient stages of working:
Example 4: 3 t a n 2 + 7 s e c − 3 = 0  3 s e c 2 +  s e c − 6 = 0     via identity followed by
1
sec= −3cos= − =  
3
Scores M1, A1, dM1, A1, A0 (only one of the two solutions given)
Example 5: 3 t a n 2 + 7 s e c − 3 = 0  6 c o s 2 −  c o s − 3 = 0
PMT
    via identities followed by
1
cos= − =  M1, A1, dM1, A1, A1
3
Question 1:
1 | ( )
3tan2+7sec−3=03 sec 2−1 +7sec−3=0 | M1
3 s e c 2 +  s e c − 6 = 0   | A1
( 3sec−)( sec+)=0sec=...cos=... | dM1
=               | A1, A1
(5)
Total 5
If a calculator is used the solution must be correct for their equation.
E.g. 6 c o s 2 −  c o s − 3 = 0  c o s = −
1
3
  
A1: One of = awrtawrt following M1, A1, dM1
Allow either answer in radians here so score for awrt 1.91 or awrt 4.37
A1: CSO. Both = awrtandawrt and no extras in the range.
If both solutions to the quadratic were given then both must be correct
………………………………………………………………………………………………………..
Methods using calculator technology and not showing all stages of working as required by question:
Example 1:
3tan2+7sec−3=0= 
No marks
Example 2:
3 t a n
3
2
s e c 2
7 s e c
s e c
3
6
0
0
3
(
s e
s
c
e
2
c
1
)
3
7 s e c 3 0    
    
+
+ 
− =
− =


−
= −
+
 = 
−
  
=
        
Scores M1, A1, dM0, A0, A0
Example 3:
3 t a n 2 + 7 s e c − 3 = 0  6 c o s 2 −  c o s − 3 = 0     via identities followed by  =              
Scores M1, A1, dM0, A0, A0
Methods using calculator technology and showing sufficient stages of working:
Example 4: 3 t a n 2 + 7 s e c − 3 = 0  3 s e c 2 +  s e c − 6 = 0     via identity followed by
1
sec= −3cos= − =  
3
Scores M1, A1, dM1, A1, A0 (only one of the two solutions given)
Example 5: 3 t a n 2 + 7 s e c − 3 = 0  6 c o s 2 −  c o s − 3 = 0
PMT
    via identities followed by
1
cos= − =  M1, A1, dM1, A1, A1
3
\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

Solutions relying entirely on calculator technology are not acceptable.\\
Solve, for $0 < \theta \leqslant 360 ^ { \circ }$, the equation

$$3 \tan ^ { 2 } \theta + 7 \sec \theta - 3 = 0$$

giving your answers to one decimal place.

\hfill \mbox{\textit{Edexcel PURE 2024 Q1}}
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