CAIE P1 2010 November — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2010
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicQuadratic trigonometric equations
TypeDirect solve: sin²/cos² substitution
DifficultyStandard +0.3 This is a standard trigonometric equation requiring substitution of sin²x = 1 - cos²x to form a quadratic in cos x, then solving the quadratic and finding angles. It's slightly above average difficulty due to the algebraic manipulation and restricted domain, but follows a well-practiced technique with no novel insight required.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
3 Solve the equation \(15 \sin ^ { 2 } x = 13 + \cos x\) for \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(15\cos^2x + \cos x - 2 = 0\)M1 \(1 - \cos^2x = \sin^2x\) & attempt simplify
\((5\cos x + 2)(3\cos x - 1) = 0\)M1 Attempt to solve 3-term quadratic for \(\cos x\)
\(113(.6)°,\ 70.5°\)A1A1 SC 1.98, 1.23 scores 1/2
[4] 
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $15\cos^2x + \cos x - 2 = 0$ | M1 | $1 - \cos^2x = \sin^2x$ & attempt simplify |
| $(5\cos x + 2)(3\cos x - 1) = 0$ | M1 | Attempt to solve 3-term quadratic for $\cos x$ |
| $113(.6)°,\ 70.5°$ | A1A1 | SC 1.98, 1.23 scores 1/2 |
| **[4]** | | |

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3 Solve the equation $15 \sin ^ { 2 } x = 13 + \cos x$ for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.

\hfill \mbox{\textit{CAIE P1 2010 Q3 [4]}}
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Q1 3 Q2 4 Q3 4 Q4 4 Q5 7 Q6 7 Q7 7 Q8 8 Q9 9 Q10 9 Q11 13