| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Mixed sin and cos linear |
| Difficulty | Moderate -0.8 This is a straightforward trigonometric equation requiring basic algebraic manipulation to isolate tan x, followed by routine use of a calculator and CAST diagram to find solutions in the given range. The algebra is simple (expand, collect terms, divide by cos x) and the solving process is standard textbook procedure with no conceptual challenges. |
| 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) \(3(2\sin x - \cos x) = 2(\sin x - 3\cos x) \rightarrow 6x - 3c = 2x - 6c \rightarrow 4x = -3c \rightarrow \tan x = -\frac{3}{4}\) | M1, A1 [2] | Expanding, collecting, use of \(t = s ÷ c\). Answer given. All correct. |
| (ii) \(x = 180 - 36.9 = 143.1°\) or \(x = 360 - 36.9 = 323.1°\) | B1, B1√ [2] | co. For 180 + first answer. |
(i) $3(2\sin x - \cos x) = 2(\sin x - 3\cos x) \rightarrow 6x - 3c = 2x - 6c \rightarrow 4x = -3c \rightarrow \tan x = -\frac{3}{4}$ | M1, A1 [2] | Expanding, collecting, use of $t = s ÷ c$. Answer given. All correct.
(ii) $x = 180 - 36.9 = 143.1°$ or $x = 360 - 36.9 = 323.1°$ | B1, B1√ [2] | co. For 180 + first answer.\begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$3(2\sin x - \cos x) = 2(\sin x - 3\cos x)$$
can be written in the form $\tan x = -\frac{4}{5}$. [2]
\item Solve the equation $3(2\sin x - \cos x) = 2(\sin x - 3\cos x)$, for $0° \leq x \leq 360°$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2010 Q1 [4]}}