| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Deduce solutions from earlier result |
| Difficulty | Moderate -0.3 Part (i) is a standard trigonometric equation requiring the identity sin²x = 1 - cos²x to convert to a quadratic in cos x, then solving - routine A-level technique. Part (ii) is a straightforward substitution using the result from (i) with x = θ/2, requiring only careful attention to the domain. This is slightly easier than average as it's a well-practiced question type with clear structure and no novel insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(4(1 - \cos^2 x) + 8\cos x - 7 = 0\) | M1 | Use \(c^2 + x^2 = 1\) |
| \(4c^2 - 8c + 3 = 0 \Rightarrow (2\cos x - 1)(2\cos x - 3) = 0\) | M1 | Attempt to solve |
| \(x = 60°\) or \(300°\) | A1A1 | [4] |
| (ii) \(\frac{1}{2}\theta = 60°\) (or \(300°\)) | M1 | Allow \(300°\) in addition |
| \(\theta = 120°\) only | A1 | [2] |
(i) $4(1 - \cos^2 x) + 8\cos x - 7 = 0$ | M1 | Use $c^2 + x^2 = 1$
$4c^2 - 8c + 3 = 0 \Rightarrow (2\cos x - 1)(2\cos x - 3) = 0$ | M1 | Attempt to solve
$x = 60°$ or $300°$ | A1A1 | [4]
(ii) $\frac{1}{2}\theta = 60°$ (or $300°$) | M1 | Allow $300°$ in addition
$\theta = 120°$ only | A1 | [2]4 (i) Solve the equation $4 \sin ^ { 2 } x + 8 \cos x - 7 = 0$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.\\
(ii) Hence find the solution of the equation $4 \sin ^ { 2 } \left( \frac { 1 } { 2 } \theta \right) + 8 \cos \left( \frac { 1 } { 2 } \theta \right) - 7 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2013 Q4 [6]}}