| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: tan/sin/cos identity manipulation |
| Difficulty | Standard +0.3 This is a standard A-level technique of converting a trigonometric equation to quadratic form using tan x = sin x/cos x, then solving. The algebraic manipulation is straightforward and the question provides scaffolding by asking to show the quadratic form first. Slightly easier than average due to the guided structure. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Replace \(\tan x\) by \(\frac{\sin x}{\cos x}\); \(1 + \frac{\sin x^2}{\cos x} = 5\cos x\) | 1 M1 | Correct formula |
| Replace \(\sin^2 x\) by \(1 - \cos^2 x\) | 1 M1 | Correct formula used in appropriate place |
| \(6\cos^2 x - \cos x - 1 (= 0)\) | 1 A1 | AG |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solution of quadratic \([c = \frac{1}{3}\) or \(\frac{1}{2}]\) | 1 M1 | Correct method seen |
| \(x = 60°\) or \(109.5°\) | 2 A1A1 | |
| Total | 3 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Replace $\tan x$ by $\frac{\sin x}{\cos x}$; $1 + \frac{\sin x^2}{\cos x} = 5\cos x$ | 1 M1 | Correct formula |
| Replace $\sin^2 x$ by $1 - \cos^2 x$ | 1 M1 | Correct formula used in appropriate place |
| $6\cos^2 x - \cos x - 1 (= 0)$ | 1 A1 | AG |
| **Total** | **3** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solution of quadratic $[c = \frac{1}{3}$ or $\frac{1}{2}]$ | 1 M1 | Correct method seen |
| $x = 60°$ or $109.5°$ | 2 A1A1 | |
| **Total** | **3** | |7 (a) Show that the equation $1 + \sin x \tan x = 5 \cos x$ can be expressed as
$$6 \cos ^ { 2 } x - \cos x - 1 = 0$$
(b) Hence solve the equation $1 + \sin x \tan x = 5 \cos x$ for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
\hfill \mbox{\textit{CAIE P1 2020 Q7 [6]}}