| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Moderate -0.3 This is a straightforward trigonometric equation requiring standard manipulation (converting tan x to sin x/cos x, using sin²x + cos²x = 1) to reach quadratic form, then solving a simple quadratic. The steps are routine and well-practiced at A-level, making it slightly easier than average but not trivial since it requires multiple algebraic steps and checking the valid range for solutions. |
| 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) \(3\sin x \tan x = 8\) Uses \(\tan = \sin \div \cos\); Uses \(\sin^2 = 1 - \cos^2\) → \(3\cos^2 x + 8\cos - 3 = 0\) | M1, M1, A1 [3] | Replaces \(t\) by s/c. Uses \(\sin^2 = 1−\cos^2\) for eqn in cosine. Answer given. |
| (ii) \((3c - 1)(c + 3) = 0\) or formula → \(\cos x = \frac{1}{3}\) as only solution. \(x = 70.5°\) or 289.5° only. | M1, A1 A1∨ [3] | Correct means of solution of quad. co. For 360° – 1st ans + no others in range. |
**(i)** $3\sin x \tan x = 8$ Uses $\tan = \sin \div \cos$; Uses $\sin^2 = 1 - \cos^2$ → $3\cos^2 x + 8\cos - 3 = 0$ | M1, M1, A1 [3] | Replaces $t$ by s/c. Uses $\sin^2 = 1−\cos^2$ for eqn in cosine. Answer given.
**(ii)** $(3c - 1)(c + 3) = 0$ or formula → $\cos x = \frac{1}{3}$ as only solution. $x = 70.5°$ or 289.5° only. | M1, A1 A1∨ [3] | Correct means of solution of quad. co. For 360° – 1st ans + no others in range.5 (i) Show that the equation $3 \sin x \tan x = 8$ can be written as $3 \cos ^ { 2 } x + 8 \cos x - 3 = 0$.\\
(ii) Hence solve the equation $3 \sin x \tan x = 8$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2007 Q5 [6]}}