| 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 cos²θ = 1 - sin²θ to convert to a quadratic, then factorising or using the quadratic formula, followed by finding angles in the given range. It's slightly easier than average as the algebraic manipulation is straightforward and the quadratic factorises neatly, making it a routine textbook exercise with clear steps. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
**Question 3:**
$4(1 - \sin^2 \theta) = 1 + \sin \theta$
at least one interim step to
$4\sin^2\theta + \sin\theta - 3 = 0$
M1
$[\theta =]$ 270°, 48.59...°, 131.4...°
A1
B1B1B1
[5]
to nearest degree or better
ignore extra values outside range; if B3 awarded, minus 1 if extra values in range.3 Show that the equation $4 \cos ^ { 2 } \theta = 1 + \sin \theta$ can be expressed as
$$4 \sin ^ { 2 } \theta + \sin \theta - 3 = 0$$
Hence solve the equation for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI C2 Q3 [5]}}