CAIE P1 2005 November — Question 1 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2005
SessionNovember
Marks4
PaperDownload PDF ↗
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. It's a routine textbook exercise with clear steps (substitute identity, rearrange to standard form, solve quadratic, find angles in given range), making it slightly easier than average but still requiring multiple techniques.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

1 Solve the equation \(3 \sin ^ { 2 } \theta - 2 \cos \theta - 3 = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

(i) \(3\sin^2\theta - 2\cos\theta - 3 = 0\)
AnswerMarks Guidance
Use of \(s^2+c^2=1\) to eliminate sine. Correct equationM1 A1
\(\cos\theta = 0, \theta = 90°\)B1 Co.
or \(\cos\theta = -\frac{2}{3}, \theta = 131.8°\)A1 Co. (to 1 d.p or more – there must be only this answer in the range 0 to 180))
(i) $3\sin^2\theta - 2\cos\theta - 3 = 0$

Use of $s^2+c^2=1$ to eliminate sine. Correct equation | M1 A1 | 

$\cos\theta = 0, \theta = 90°$ | B1 | Co.

or $\cos\theta = -\frac{2}{3}, \theta = 131.8°$ | A1 | Co. (to 1 d.p or more – there must be only this answer in the range 0 to 180)) | [4]
1 Solve the equation $3 \sin ^ { 2 } \theta - 2 \cos \theta - 3 = 0$, for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P1 2005 Q1 [4]}}