| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Convert equation to quadratic form |
| Difficulty | Standard +0.3 This is a standard trigonometric equation requiring routine manipulation (converting tan x to sin x/cos x, multiplying through by cos x, using sin²x + cos²x = 1) followed by solving a quadratic in cos x. The steps are methodical and commonly practiced, making it slightly easier than average but not trivial due to the multi-step algebraic manipulation 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) \(2\sin x \tan x + 3 = 0\) | ||
| \(2\sin x \frac{\sin x}{\cos x} + 3 = 0\) | M1 | For using \(\tan = \sin \div \cos\) |
| \(2\frac{(1-\cos^2 x)}{\cos x} + 3 = 0\) | M1 | For using \(\sin^2 + \cos^2 = 1\) and everything correct |
| \(\to 2\cos^2 x - 3\cos x - 2 = 0\) | [2] | Answer given – check. |
| (ii) \(2\cos^2 x - 3\cos x - 2 = 0\) | M1 | Solution of quadratic. |
| \(\to \cos x = -\frac{1}{2}\) or \(2\) | A1 B1√ | co. √ for \(360 –\) his answer. |
| \(x = 120°\) or \(240°\) | [3] |
(i) $2\sin x \tan x + 3 = 0$ | |
$2\sin x \frac{\sin x}{\cos x} + 3 = 0$ | M1 | For using $\tan = \sin \div \cos$
$2\frac{(1-\cos^2 x)}{\cos x} + 3 = 0$ | M1 | For using $\sin^2 + \cos^2 = 1$ and everything correct
$\to 2\cos^2 x - 3\cos x - 2 = 0$ | [2] | Answer given – check.
(ii) $2\cos^2 x - 3\cos x - 2 = 0$ | M1 | Solution of quadratic.
$\to \cos x = -\frac{1}{2}$ or $2$ | A1 B1√ | co. √ for $360 –$ his answer.
$x = 120°$ or $240°$ | [3] |
---4 (i) Show that the equation $2 \sin x \tan x + 3 = 0$ can be expressed as $2 \cos ^ { 2 } x - 3 \cos x - 2 = 0$.\\
(ii) Solve the equation $2 \sin x \tan x + 3 = 0$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P1 2010 Q4 [5]}}