| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Moderate -0.8 This is a straightforward textbook exercise requiring standard techniques: using the identity cos²θ = 1 - sin²θ to convert to a quadratic, then factorising and solving. The factorisation is simple (2sin θ - 1)(sin θ - 3) = 0, and finding angles in the given range is routine. Below average difficulty as it's purely procedural with no problem-solving or insight required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(2(1-\sin^2\theta) + 7\sin\theta = 5\) leading to \(2\sin^2\theta - 7\sin\theta + 3 = 0\) | B1 | Use of \(\cos^2\theta = 1 - \sin^2\theta\) |
| (ii) \((2\sin\theta - 1)(\sin\theta - 3) = 0\) | M1 A1 | |
| \(\sin\theta = \frac{1}{2}\) (\(\sin\theta = 3\) rejected) | M1 | |
| \(\theta = 30°, 150°\) | A1 A1 |
## Question 8:
**(i)** $2(1-\sin^2\theta) + 7\sin\theta = 5$ leading to $2\sin^2\theta - 7\sin\theta + 3 = 0$ | B1 | Use of $\cos^2\theta = 1 - \sin^2\theta$
**(ii)** $(2\sin\theta - 1)(\sin\theta - 3) = 0$ | M1 A1 |
$\sin\theta = \frac{1}{2}$ ($\sin\theta = 3$ rejected) | M1 |
$\theta = 30°, 150°$ | A1 A1 |
---8 (i) Show that the equation $2 \cos ^ { 2 } \theta + 7 \sin \theta = 5$ may be written in the form
$$2 \sin ^ { 2 } \theta - 7 \sin \theta + 3 = 0$$
(ii) By factorising this quadratic equation, solve the equation for values of $\theta$ between $0 ^ { \circ }$ and $180 ^ { \circ }$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI C2 2007 Q8 [5]}}