| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Two-part show and solve with cos²θ quartic form |
| Difficulty | Standard +0.3 This is a standard two-part trigonometric equation requiring routine algebraic manipulation (converting tan and sin to cos using identities) followed by solving a quadratic in cos²x. The techniques are well-practiced at this level, though the multi-step nature and need to handle a quartic form elevates it slightly above average difficulty. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3\sin^2 x - 3\sin^2 x \cos^2 x - 4\cos^2 x [= 0]\) | M1 | Replace \(\tan^2 x\) with \(\frac{\sin^2 x}{\cos^2 x}\) and multiply by \(\cos^2 x\) |
| \(3(1 - \cos^2 x) - 3(1 - \cos^2 x)\cos^2 x - 4\cos^2 x [= 0]\) | M1 | Replace \(\sin^2 x\) by \(1 - \cos^2 x\) twice |
| \(3\cos^4 x - 10\cos^2 x + 3 = 0\) or \(-3\cos^4 x + 10\cos^2 x - 3 = 0\) | A1 | Or multiple of these equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3\cos^2 x - 1)(\cos^2 x - 3) [= 0]\) | M1 | OE, using their equation in the given form. Allow unusual notation if meaning is clear |
| \(\cos x = [\pm]\frac{1}{\sqrt{3}}\) | A1 | SOI Answer only SC B1 |
| \(54.7°\) | A1 | |
| \(125.3°\) | A1 FT | Only other answer and must be from correct factorisation for A1. FT for \(180°\) − their first answer. Answers only SC B1, SC B1 FT |
## Question 4:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3\sin^2 x - 3\sin^2 x \cos^2 x - 4\cos^2 x [= 0]$ | M1 | Replace $\tan^2 x$ with $\frac{\sin^2 x}{\cos^2 x}$ and multiply by $\cos^2 x$ |
| $3(1 - \cos^2 x) - 3(1 - \cos^2 x)\cos^2 x - 4\cos^2 x [= 0]$ | M1 | Replace $\sin^2 x$ by $1 - \cos^2 x$ twice |
| $3\cos^4 x - 10\cos^2 x + 3 = 0$ or $-3\cos^4 x + 10\cos^2 x - 3 = 0$ | A1 | Or multiple of these equations |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3\cos^2 x - 1)(\cos^2 x - 3) [= 0]$ | M1 | OE, using their equation in the given form. Allow unusual notation if meaning is clear |
| $\cos x = [\pm]\frac{1}{\sqrt{3}}$ | A1 | SOI Answer only SC B1 |
| $54.7°$ | A1 | |
| $125.3°$ | A1 FT | Only other answer and must be from correct factorisation for A1. FT for $180°$ − their first answer. Answers only SC B1, SC B1 FT |
---4
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$3 \tan ^ { 2 } x - 3 \sin ^ { 2 } x - 4 = 0$$
may be expressed in the form $a \cos ^ { 4 } x + b \cos ^ { 2 } x + c = 0$, where $a , b$ and $c$ are constants to be found.
\item Hence solve the equation $3 \tan ^ { 2 } x - 3 \sin ^ { 2 } x - 4 = 0$ for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q4 [7]}}