Specimen 2020 Paper 1 Q20
20
(a) Show that \(\quad 2\cos^2\theta \equiv 2 - 2\sin^2\theta\) [1 mark]
(b) Hence, solve \(\quad 2\cos^2\theta + 3\sin\theta = 3 \qquad\) for \(\ 0 \lt \theta \lt 180^\circ\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2(1 - \sin^2\theta) = 2 - 2\sin^2\theta\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| \(2\sin^2\theta - 3\sin\theta + 1\ \ (= 0)\) | M1 | |
| \((\sin\theta - 1)(2\sin\theta - 1)\ \ (= 0)\) | M1dep | |
| \(\sin\theta = 1\) or \(\sin\theta = \dfrac{1}{2}\) | M1dep | |
| 30° and 90° and 150° | A1 |