June 2025 Paper 2 Q12
12 In this question you must show detailed reasoning
Solve the equation \(3\sec^2\theta + 2\tan\theta - 4 = 0\) for \(-180^\circ \lt \theta \lt 180^\circ\). [6]
| Scheme | Marks | AO |
|---|---|---|
| DR \(3(1 + \tan^2\theta) + 2\tan\theta - 4\ [=0]\) | B1 | 3.1a |
| \(3\tan^2\theta + 2\tan\theta - 1 = 0\) | M1 | 1.1 |
| \((3\tan\theta - 1)(\tan\theta + 1)\ [=0]\) | B1 | 2.1 |
| \(\tan\theta = -1,\ \tan\theta = \frac{1}{3}\) | A1 | 1.1 |
| \(-45^\circ,\ 135^\circ,\ 18.4^\circ\) to \(18.435^\circ,\ -161.6^\circ\) to \(-161.565^\circ\) or \(-162^\circ\) | A1 | 1.1 |
| A1 | 3.2a | |
| [6] |
Notes
DR: This question included the instruction: In this question you must show detailed reasoning.
B1: correct substitution of \(\sec^2\theta = 1 + \tan^2\theta\) in equation; may be implied by \(3 + 3\tan^2\theta + 2\tan\theta - 4\ [=0]\)
M1: 3 term quadratic in \(\tan\theta\) obtained following attempt at use of Pythagoras; may be implied if equation correct, otherwise substitution must be seen; allow one coefficient error
B1: value(s) for \(\tan\theta\) obtained from their quadratic by valid method; must see correct factorisation or correct use of formula oe
A1: both correct
A1: any two values correct; dependent on award of previous A1
A1: all four values correct with no extras in the range; ignore values outside the range
if A0A0 allow SC1 for \(-\frac{\pi}{4}, \frac{3\pi}{4}\), 0.32175 to 0.322, \(-2.8198\) to \(-2.82\)
NB B0M1B0A1A1A1 is possible
Alternative method 1
| Scheme | Marks |
|---|---|
| \(3 + 2\sin\theta\cos\theta - 4\cos^2\theta = 0\) | |
| \(3 + \sin 2\theta - 4\left(\frac{1}{2}(\cos 2\theta + 1)\right)\ [=0]\) | B1 |
| \(\sqrt{5}\sin(2\theta - 63.4) = -1\) or \(\sqrt{5}\cos(2\theta + 63.4) = 1\) | M1 |
| \(\sin(2\theta - 63.4) = -\frac{1}{\sqrt{5}}\) or \(\cos(2\theta + 26.6) = \frac{1}{\sqrt{5}}\) | M1 A1 |
| \(-45^\circ,\ 135^\circ,\ 18.4^\circ\) to \(18.435^\circ,\ -161.6^\circ\) to \(-161.565^\circ\) or \(-162^\circ\) | A1 |
| A1 |
from multiplying through by \(\cos^2\theta\)
B1: substitution of correct double angle formulae in correct equation
M1: \(R\sin(2\theta \pm \alpha) = k\) or \(R\cos(2\theta \pm \alpha) = k\)
M1 A1: value for \(\sin(2\theta - 63.4) = -\frac{1}{\sqrt{5}}\) or \(\cos(2\theta + 26.6) = \frac{1}{\sqrt{5}}\) obtained by valid method; A1 if all correct
A1: any two values correct; dependent on award of previous A1
A1: all four values correct with no extras in the range; ignore values outside the range
if A0A0 allow SC1 for \(-\frac{\pi}{4}, \frac{3\pi}{4}\), 0.32175 to 0.322, \(-2.8198\) to \(-2.82\)
Alternative method 2
| Scheme | Marks |
|---|---|
| \((2\sin\theta\cos\theta)^2 = (4\cos^2\theta - 3)^2\) | |
| \(4\sin^2\theta - 4\sin^4\theta = (1 - 4\sin^2\theta)^2\) oe or \(4\cos^2\theta - 4\cos^4\theta = (4\cos^2\theta - 3)^2\) oe | B1 |
| \(20\sin^4\theta - 12\sin^2\theta + 1 = 0\) or \(20\cos^4\theta - 28\cos^2\theta + 9 = 0\) | M1 |
| \(\sin^2\theta = 0.5,\ 0.1\) or \(\cos^2\theta = 0.5,\ 0.9\) | B1 |
| A1 | |
| \(-45^\circ,\ 135^\circ,\ 18.4^\circ\) to \(18.435^\circ,\ -161.6^\circ\) to \(-161.565^\circ\) or \(-162^\circ\) | A1 |
| A1 |
from multiplying through by \(\cos^2\theta\) and squaring both sides
B1: correct substitution of \(\cos^2\theta = 1 - \sin^2\theta\) or \(\sin^2\theta = 1 - \cos^2\theta\) in correct equation
M1: 3 term quadratic in \(\sin^2\theta\) or \(\cos^2\theta\) obtained following use of Pythagoras – may be implied if equation correct, otherwise substitution must be seen; allow one coefficient error
B1: values for \(\sin^2\theta\) or \(\cos^2\theta\) obtained from their quadratic by valid method; must see correct factorisation or correct use of formula
A1: both values of \(\sin^2\theta\) or \(\cos^2\theta\) correct
A1: any two values correct; dependent on award of previous A1
A1: all four values correct with no extras in the range; ignore values outside the range
if A0A0 allow SC1 for \(-\frac{\pi}{4}, \frac{3\pi}{4}\), 0.32175 to 0.322, \(-2.8198\) to \(-2.82\)
NB B0M1B0A1A1A1 is possible