CAIE P1 2022 November — Question 1 3 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2022
SessionNovember
Marks3
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Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeDirect solve: sin²/cos² substitution
DifficultyModerate -0.3 This is a standard trigonometric equation requiring substitution of sin²θ = 1 - cos²θ to form a quadratic in cos θ, then solving. It's routine for P1 level with straightforward algebraic manipulation and a restricted domain, making it slightly easier than average but not trivial.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
1 Solve the equation \(8 \sin ^ { 2 } \theta + 6 \cos \theta + 1 = 0\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(8(1-\cos^2\theta) + 6\cos\theta + 1 = 0\)M1 Expect \(8\cos^2\theta - 6\cos\theta - 9 = 0\)
\((4\cos\theta + 3)(2\cos\theta - 3) = 0\)A1 Factors or formula or completing square must be shown
\(\cos\theta = -0.75 \rightarrow \theta =\) 138.6° onlyA1 AWRT, ignore solutions outside the given range, answer in radians A0
Total: 3 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $8(1-\cos^2\theta) + 6\cos\theta + 1 = 0$ | M1 | Expect $8\cos^2\theta - 6\cos\theta - 9 = 0$ |
| $(4\cos\theta + 3)(2\cos\theta - 3) = 0$ | A1 | Factors or formula or completing square must be shown |
| $\cos\theta = -0.75 \rightarrow \theta =$ 138.6° only | A1 | AWRT, ignore solutions outside the given range, answer in radians A0 |
| **Total: 3** | | |

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1 Solve the equation $8 \sin ^ { 2 } \theta + 6 \cos \theta + 1 = 0$ for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.\\

\hfill \mbox{\textit{CAIE P1 2022 Q1 [3]}}
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Q1 3 Q2 6 Q3 6 Q4 5 Q5 6 Q6 5 Q7 7 Q8 7 Q9 9 Q10 10 Q11 11