CAIE P1 2021 March — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2021
SessionMarch
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeRational trig expressions
DifficultyStandard +0.3 This requires converting tan to sin/cos, algebraic manipulation to reach a standard form, and solving a basic quadratic in cos θ. While it involves multiple steps, the techniques are standard for P1 level with no novel insight required, making it slightly easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
3 Solve the equation \(\frac { \tan \theta + 2 \sin \theta } { \tan \theta - 2 \sin \theta } = 3\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(\tan\theta + 2\sin\theta = 3\tan\theta - 6\sin\theta\) leading to \(2\tan\theta - 8\sin\theta [= 0]\)M1 OE
\(2\sin\theta - 8\sin\theta\cos\theta (= 0)\) leading to \([2]\sin\theta(1 - 4\cos\theta)[= 0]\)M1
\(\cos\theta = \frac{1}{4}\)A1 Ignore \(\sin\theta = 0\)
\(\theta = 75.5°\) onlyA1 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\theta + 2\sin\theta = 3\tan\theta - 6\sin\theta$ leading to $2\tan\theta - 8\sin\theta [= 0]$ | M1 | OE |
| $2\sin\theta - 8\sin\theta\cos\theta (= 0)$ leading to $[2]\sin\theta(1 - 4\cos\theta)[= 0]$ | M1 | |
| $\cos\theta = \frac{1}{4}$ | A1 | Ignore $\sin\theta = 0$ |
| $\theta = 75.5°$ only | A1 | |

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3 Solve the equation $\frac { \tan \theta + 2 \sin \theta } { \tan \theta - 2 \sin \theta } = 3$ for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.\\

\hfill \mbox{\textit{CAIE P1 2021 Q3 [4]}}
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