| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Deduce related solution |
| Difficulty | Moderate -0.3 Part (i) is a standard trigonometric equation requiring the identity sin²x = 1 - cos²x to convert to a quadratic in cos x, then solving and rejecting an impossible solution (cos x > 1). Part (ii) applies the result via substitution and solving for θ using inverse cosine. This is routine bookwork for P1 level with no novel insight required, making it slightly easier than average but not trivial due to the multi-step nature and need for careful angle 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 |
|---|---|---|
| \(\cos x = -\frac{2}{3}\) | M1, M1, A1 [3] | Use of \(c^2 + s^2 = 1\); Factorising, formula or completing the square needed; AG Ignore \(\cos x = -2\) also offered; SC B1 if -2/3 and -2 seen |
| Answer | Marks |
|---|---|
| \(\theta = 158.2\) | M1A1, M1, A1 [4] |
**(i)** $3\cos^2x + 8\cos x + 4 = 0$
$(3\cos x + 2)(\cos x + 2) = 0$
$\cos x = -\frac{2}{3}$ | M1, M1, A1 [3] | Use of $c^2 + s^2 = 1$; Factorising, formula or completing the square needed; AG Ignore $\cos x = -2$ also offered; SC B1 if -2/3 and -2 seen
**(ii)** $\cos(\theta + 70) = -\frac{2}{3}$, $\theta = 61.8$
$\theta + 70 = 131.8$ (or 228.2)
$\theta = 158.2$ | M1A1, M1, A1 [4]
---\begin{enumerate}[label=(\roman*)]
\item Given that
$$3\sin^2 x - 8\cos x - 7 = 0,$$
show that, for real values of $x$,
$$\cos x = -\frac{2}{3}.$$ [3]
\item Hence solve the equation
$$3\sin^2(\theta + 70°) - 8\cos(\theta + 70°) - 7 = 0$$
for $0° \leqslant \theta \leqslant 180°$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q5 [7]}}