| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Moderate -0.3 This is a standard C2 trigonometric equation requiring the identity sin²θ + cos²θ = 1 to convert to a quadratic in cos θ, then solving a straightforward quadratic. The steps are routine and well-practiced at this level, making it slightly easier than average but not trivial since it requires multiple techniques (identity, quadratic formula, inverse trig). |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\sin^2\theta + 3\cos\theta = 2 - 2\cos^2\theta + 3\cos\theta\) | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\sin^2\theta + 3\cos\theta = 3 \Rightarrow 2 - 2\cos^2\theta + 3\cos\theta = 3\) | B1 | |
| \(\Rightarrow 2\cos^2\theta - 3\cos\theta + 1 = 0\) | M1 | |
| \(\Rightarrow (2\cos\theta - 1)(\cos\theta - 1) = 0 \Rightarrow \cos\theta = 1, \frac{1}{2}\) | A1 | For \(0°, 60°, 360°\) |
| \(\Rightarrow \theta = 0°, 60°, 300°, 360°\) | A1 | For \(300°\) [5] |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\sin^2\theta + 3\cos\theta = 2 - 2\cos^2\theta + 3\cos\theta$ | B1 | **[1]** |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\sin^2\theta + 3\cos\theta = 3 \Rightarrow 2 - 2\cos^2\theta + 3\cos\theta = 3$ | B1 | |
| $\Rightarrow 2\cos^2\theta - 3\cos\theta + 1 = 0$ | M1 | |
| $\Rightarrow (2\cos\theta - 1)(\cos\theta - 1) = 0 \Rightarrow \cos\theta = 1, \frac{1}{2}$ | A1 | For $0°, 60°, 360°$ |
| $\Rightarrow \theta = 0°, 60°, 300°, 360°$ | A1 | For $300°$ **[5]** |
---5 (i) Express $2 \sin ^ { 2 } \theta + 3 \cos \theta$ as a quadratic function of $\cos \theta$.\\
(ii) Hence solve the equation $2 \sin ^ { 2 } \theta + 3 \cos \theta = 3$, giving all values of $\theta$ correct to the nearest degree in the range $0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI C2 Q5 [5]}}