June 2024 Paper 3 Q6
6 In this question you must show detailed reasoning.
Solve the equation \(\tan x - 3\cot x = 2\) for values of \(x\) in the interval \(0^\circ \leqslant x \leqslant 360^\circ\). [5]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\tan^2 x - 3 = 2\tan x\) | M1^ | 3.1a |
| \(\tan^2 x - 2\tan x - 3 = 0\) | M1^ | 1.1 |
| \((\tan x - 3)(\tan x + 1) = 0\) \(\tan x = 3\), \(\tan x = -1\) | M1* | 1.1 |
| \(71.6^\circ\), \(251.6^\circ\), \(135^\circ\), \(315^\circ\) | A1^ A1* | 1.1 1.1 |
| [5] |
Notes
M1^: Multiplying through by \(\tan x\)
M1^: Get all 3 terms of quadratic on one side and zero on the other
M1*: Solve 3 term quadratic to get both values of \(\tan x\)
Must see either use of the formula, factorisation or completing the square. Allow 1 error in method
Condone missing ‘= 0’
A1^: for any two roots correct (dep. on M2^)
A1*: for all roots with no additional solutions (dep. on M1*)
Accept 72 and 252
Alternative solution
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\sin x}{\cos x} - 3\dfrac{\cos x}{\sin x} = 2\) | M1^ | 3.1a |
| \(\sin^2 x - 2\sin x\cos x - 3\cos^2 x = 0\) | M1^ | 1.1 |
| \((\sin x - 3\cos x)(\sin x + \cos x) = 0\) \(\tan x = 3\), \(\tan x = -1\) | M1* | 1.1 |
| \(71.6^\circ\), \(251.6^\circ\), \(135^\circ\), \(315^\circ\) | A1^ A1* | 1.1 1.1 |
M1^: Use of \(\tan x = \dfrac{\sin x}{\cos x}\)
M1^: Get all 3 terms on one side and zero on the other
M1*: Factorise to get both values of \(\tan x\)
Must see factorisation. Allow 1 error in method
Condone missing =0
A1^: for any two roots correct (dep. on M2^)
A1*: for all roots with no additional solutions (dep. on M1*)
Accept 72 and 252
Additional guidance
This is another DR question. There are 2 solutions provided.
The first method involves multiplying through by tan x (M1) and the second method involves replacing tan x with sin x/cos x and cot x with cos x/sin x (M1). Both methods then involve getting an equation with 3 terms on one side and zero on the other (M1) and solving their 3-term equation AND get both values of tan x (M1). A correctly solved equation implies any missing M marks. They must solve the equation by a valid method (factorisation, formula or completing the square) to get the M1 here.
The A marks are for giving the angles – A1 for any two correct (subject to at least M2) and A2 for all the roots with no additional solutions (subject to M3).
It might be quite common to give M1 M1 M0 A1 A0 because they fail to solve the equation and just write down answers. Note we are allowing 72°/ 252° and also over specified answers. e.g. 71.565°/ 251.565°.