| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trigonometric equations in context |
| Type | Reduce to quadratic in trig |
| Difficulty | Moderate -0.3 Both parts are standard C2 trigonometric equations requiring routine techniques: (i) uses the Pythagorean identity to convert to a quadratic in cos x, (ii) divides to get tan 2x = -1. The restricted domain and straightforward algebraic manipulation make these slightly easier than average A-level questions, though they do require knowing standard identities. |
| Spec | 1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos x = -1 \Rightarrow x = 180°\) | M1, M1, A1, A1 | For transforming to a quadratic in \(\cos x\). For solution of a quadratic in \(\cos x\). For correct answer \(60°\). For correct answer \(180°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(x = 67.5\) or \(157.5\) | M1, M1, A1, A1 | For transforming to an equation of form \(\tan 2x = k\). For correct solution method, i.e. inverse tan followed by division by 2. For correct value 67.5. For correct value 157.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(x = 67.5\) or \(157.5\) | M1, M1, A1, A1 | Obtain linear equation in \(\cos 2x\) or \(\sin 2x\). Use correct solution method. For correct value 67.5. For correct value 157.5 |
**(i)** $\sin^2 x = 1 - \cos^2 x \Rightarrow 2\cos^2 x + \cos x - 1 = 0$
Hence $(2\cos x - 1)(\cos x + 1) = 0$
$\cos x = \frac{1}{2} \Rightarrow x = 60°$
$\cos x = -1 \Rightarrow x = 180°$ | M1, M1, A1, A1 | For transforming to a quadratic in $\cos x$. For solution of a quadratic in $\cos x$. For correct answer $60°$. For correct answer $180°$ | 4 marks
[Max 3 out of 4 if any extra answers present in range, or in radians]
**SR answer only is B1, B1; justification – ie graph or substitution is B2,**
**(ii)** $\tan 2x = -1 \Rightarrow 2x = 135$ or $315$
Hence $x = 67.5$ or $157.5$ | M1, M1, A1, A1 | For transforming to an equation of form $\tan 2x = k$. For correct solution method, i.e. inverse tan followed by division by 2. For correct value 67.5. For correct value 157.5 | 4 marks
**SR answer only is B1, B1; justification – ie graph or substitution is B2, B2**
**OR**
$\sin^2 2x = \cos^2 2x$
$2\sin^2 2x = 1$ and $2\cos^2 2x = 1$
$\sin 2x = \pm\frac{1}{\sqrt{2}}$ and $\cos 2x = \pm\frac{1}{\sqrt{2}}$
Hence $x = 67.5$ or $157.5$ | M1, M1, A1, A1 | Obtain linear equation in $\cos 2x$ or $\sin 2x$. Use correct solution method. For correct value 67.5. For correct value 157.5 | 8 marks
---5 Solve each of the following equations, for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
(i) $2 \sin ^ { 2 } x = 1 + \cos x$.\\
(ii) $\sin 2 x = - \cos 2 x$.
\hfill \mbox{\textit{OCR C2 2006 Q5 [8]}}