CAIE P1 2016 June — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionJune
Marks4
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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 the identity sin²θ = 1 - cos²θ to convert to a quadratic in cos θ, then solving the resulting quadratic and finding angles. It's slightly easier than average as it's a routine textbook exercise with a clear method, though it does require multiple steps (substitution, quadratic formula/factoring, and inverse trig for multiple solutions in the given range).
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
2 Solve the equation \(3 \sin ^ { 2 } \theta = 4 \cos \theta - 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(3\sin^2\theta = 4\cos\theta - 1\); uses \(s^2 + c^2 = 1\) giving \(3c^2 + 4c - 4 = 0\)M1 A1 Equation in \(\cos\theta\) only, all terms on one side
\(c = \frac{2}{3}\) or \(-2\)
\(\theta = 48.2°\) or \(311.8°\); 0.841, 5.44 rads (A1 only); \((0.268\pi, 1.73\pi)\)A1 A1\(\checkmark\) For \(360°\) − 1st answer 
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3\sin^2\theta = 4\cos\theta - 1$; uses $s^2 + c^2 = 1$ giving $3c^2 + 4c - 4 = 0$ | **M1 A1** | Equation in $\cos\theta$ only, all terms on one side |
| $c = \frac{2}{3}$ or $-2$ | | |
| $\theta = 48.2°$ or $311.8°$; 0.841, 5.44 rads (**A1** only); $(0.268\pi, 1.73\pi)$ | **A1 A1$\checkmark$** | For $360°$ − 1st answer |
2 Solve the equation $3 \sin ^ { 2 } \theta = 4 \cos \theta - 1$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P1 2016 Q2 [4]}}
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