| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 6 |
| 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 reciprocal trig identity question requiring routine algebraic manipulation (converting to sin/cos, finding common denominator, using sin²x + cos²x = 1) followed by solving a quadratic in cos x. While multi-step, it follows a predictable pattern with no novel insight required, making it slightly easier than average. |
| 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 |
| \(4\sin^2 x + 5\cos x + 2\ [=0]\) | *M1 | Multiply by \(\sin x\) (or writing as single fraction) and using \(\tan x = \frac{\sin x}{\cos x}\) |
| \(4(1-\cos^2 x) + 5\cos x + 2\ [=0]\) | DM1 | Correctly obtaining a quadratic in \(\cos x\) (allow sign errors) |
| \(4\cos^2 x - 5\cos x - 6 = 0\) | A1 | Condone missing \(x\). Must be \(= 0\) unless \(0\) appears on RHS earlier |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((4\cos x + 3)(\cos x - 2)\ [=0]\) | M1 | Or use of formula or completing the square |
| \(138.6°,\ 221.4°\) | A1 B1 FT | FT on \(360°\) – 1st solution from quadratic in \(\cos x\). Use of radians (2.42) A0 but allow B1 FT for \(2\pi\): 1st solution if use of radians is clear. SC If M0 scored SC B1 B1 for correct final answer(s). If extra incorrect solutions in range \(0 \to 360°\) are given award A1 B0 |
| Total: 3 |
## Question 5:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\sin^2 x + 5\cos x + 2\ [=0]$ | *M1 | Multiply by $\sin x$ (or writing as single fraction) and using $\tan x = \frac{\sin x}{\cos x}$ |
| $4(1-\cos^2 x) + 5\cos x + 2\ [=0]$ | DM1 | Correctly obtaining a quadratic in $\cos x$ (allow sign errors) |
| $4\cos^2 x - 5\cos x - 6 = 0$ | A1 | Condone missing $x$. Must be $= 0$ unless $0$ appears on RHS earlier |
| **Total: 3** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(4\cos x + 3)(\cos x - 2)\ [=0]$ | M1 | Or use of formula or completing the square |
| $138.6°,\ 221.4°$ | A1 B1 FT | FT on $360°$ – 1st solution from quadratic in $\cos x$. Use of radians (2.42) A0 but allow B1 FT for $2\pi$: 1st solution if use of radians is clear. SC If M0 scored SC B1 B1 for correct final answer(s). If extra incorrect solutions in range $0 \to 360°$ are given award A1 B0 |
| **Total: 3** | | |
---5
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$4 \sin x + \frac { 5 } { \tan x } + \frac { 2 } { \sin x } = 0$$
may be expressed in the form $a \cos ^ { 2 } x + b \cos x + c = 0$, where $a , b$ and $c$ are integers to be found.
\item Hence solve the equation $4 \sin x + \frac { 5 } { \tan x } + \frac { 2 } { \sin x } = 0$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q5 [6]}}