| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Convert to quadratic in sin/cos |
| Difficulty | Moderate -0.3 This is a standard two-part trigonometric equation question requiring routine algebraic 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 techniques are straightforward for P1 level with no novel insight required, making it slightly easier than average but not trivial due to the multi-step algebraic manipulation. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow 6c^2 - c - 1 (= 0)\) | M1, M1, A1 [3] | Correct formula; Correct formula used in appropriate place; AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\rightarrow x = 60°\) or \(109.5°\) | M1, A1, A1 [3] | Correct method; co co |
$1 + \sin x \tan x = 5\cos x$
**(i)** Replaces s by s/c
$1 + \frac{s^2}{c} = 5c$
Replace $s^2$ by $1 - c^2$
$\rightarrow 6c^2 - c - 1 (= 0)$ | M1, M1, A1 [3] | Correct formula; Correct formula used in appropriate place; AG
**(ii)** Soln of quadratic $\rightarrow (c = -\frac{1}{2}$ or $\frac{1}{3})$
$\rightarrow x = 60°$ or $109.5°$ | M1, A1, A1 [3] | Correct method; co co\begin{enumerate}[label=(\roman*)]
\item Show that the equation $1 + \sin x \tan x = 5 \cos x$ can be expressed as
$$6 \cos^2 x - \cos x - 1 = 0.$$ [3]
\item Hence solve the equation $1 + \sin x \tan x = 5 \cos x$ for $0° \leqslant x \leqslant 180°$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2014 Q5 [6]}}