| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard trigonometric equations |
| Type | Multiple independent equations — all direct solve |
| Difficulty | Standard +0.3 Part (a) requires straightforward manipulation of inverse trig functions and exact value recall. Part (b) involves a standard double angle substitution leading to a quadratic in sin θ, then finding solutions in a given range. Both are routine C3 techniques with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\arctan(x-2) = -\frac{\pi}{3}\) | M1 | |
| \(x - 2 = \tan(-\frac{\pi}{3}) = -\sqrt{3}\) | M1 | |
| \(x = 2 - \sqrt{3}\) | A1 | |
| (b) \(1 - 2\sin^2\theta - \sin\theta - 1 = 0\) | M1 | |
| \(2\sin^2\theta + \sin\theta = 0\) | M1 | |
| \(\sin\theta(2\sin\theta + 1) = 0\) | M1 | |
| \(\sin\theta = 0\) or \(-\frac{1}{2}\) | A1 | |
| \(\theta = 0\) or \(-\frac{\pi}{6}, -\pi + \frac{\pi}{6}\) | A2 | (8) |
| \(\theta = -\frac{5\pi}{6}, -\frac{\pi}{6}, 0\) |
(a) $\arctan(x-2) = -\frac{\pi}{3}$ | M1 |
$x - 2 = \tan(-\frac{\pi}{3}) = -\sqrt{3}$ | M1 |
$x = 2 - \sqrt{3}$ | A1 |
(b) $1 - 2\sin^2\theta - \sin\theta - 1 = 0$ | M1 |
$2\sin^2\theta + \sin\theta = 0$ | M1 |
$\sin\theta(2\sin\theta + 1) = 0$ | M1 |
$\sin\theta = 0$ or $-\frac{1}{2}$ | A1 |
$\theta = 0$ or $-\frac{\pi}{6}, -\pi + \frac{\pi}{6}$ | A2 | **(8)**
$\theta = -\frac{5\pi}{6}, -\frac{\pi}{6}, 0$ |\begin{enumerate}[label=(\alph*)]
\item Find the exact value of $x$ such that
$$3 \arctan (x - 2) + \pi = 0.$$ [3]
\item Solve, for $-\pi < \theta < \pi$, the equation
$$\cos 2\theta - \sin \theta - 1 = 0,$$
giving your answers in terms of $\pi$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q1 [8]}}