CAIE P1 2012 November — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
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 sin²x = 1 - cos²x to convert to a quadratic in cos x, then solving the quadratic and finding angles. It's slightly easier than average as it's a routine textbook exercise with a well-known technique, though it does require multiple steps (substitution, quadratic formula, inverse trig for two 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

3 Solve the equation \(7 \cos x + 5 = 2 \sin ^ { 2 } x\), for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

AnswerMarks Guidance
\(7\cos x + 5 = 2(1-\cos^2 x)\); \((2\cos x + 1)(\cos x + 3) = 0\); \(\cos x = -0.5\); \(x = 120°, 240°\)M1, A1, A1, A1/\(\checkmark\) Use of \(c^2 + s^2 = 1\); ft for 360 -1st solution [4]
$7\cos x + 5 = 2(1-\cos^2 x)$; $(2\cos x + 1)(\cos x + 3) = 0$; $\cos x = -0.5$; $x = 120°, 240°$ | M1, A1, A1, A1/$\checkmark$ | Use of $c^2 + s^2 = 1$; ft for 360 -1st solution [4]
3 Solve the equation $7 \cos x + 5 = 2 \sin ^ { 2 } x$, for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P1 2012 Q3 [4]}}