CAIE P2 2014 June — Question 2 4 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2014
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeSolve equation with tan2x or mixed tan/double angle
DifficultyStandard +0.3 This question requires applying the double angle formula (sin 2θ = 2sin θ cos θ) and tan θ = sin θ/cos θ, then algebraic manipulation to reach a quadratic in sin θ or cos θ. While it involves multiple steps and careful handling of the domain restriction, it's a fairly standard application of double angle formulae without requiring novel insight—slightly easier than average due to being a routine technique-based problem.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

2 Solve the equation \(3 \sin 2 \theta \tan \theta = 2\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

AnswerMarks
Use \(\sin 2\theta = 2\sin\theta\cos\theta\)B1
Simplify to obtain form \(c_1\sin^2\theta = c_2\) or equivalentM1
Find at least one value of \(\theta\) from equation of form \(\sin\theta = k\)M1
Obtain \(35.3°\) and \(144.7°\)A1 [4]
Use $\sin 2\theta = 2\sin\theta\cos\theta$ | B1 |
Simplify to obtain form $c_1\sin^2\theta = c_2$ or equivalent | M1 |
Find at least one value of $\theta$ from equation of form $\sin\theta = k$ | M1 |
Obtain $35.3°$ and $144.7°$ | A1 [4] |
2 Solve the equation $3 \sin 2 \theta \tan \theta = 2$ for $0 ^ { \circ } < \theta < 180 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P2 2014 Q2 [4]}}