June 2025 Paper 1 Q9
9
Solve the equation \(\dfrac{3\cos 2x + 4\sin 2x + 6}{3\cos 2x + 4\sin 2x - 1} = 4\) for \(0^\circ \lt x \lt 360^\circ\). [7]
| Scheme | Marks | AO |
|---|---|---|
| \(R^2 = 25\) \(R = 5\) | M1 | 1.1 |
| \(R\cos\alpha = 3\), \(R\sin\alpha = 4\) hence \(\tan\alpha = \frac{4}{3}\) | M1 | 1.1a |
| \(5\cos(2x - 53.1)\) | A1 | 1.1 |
| [3] |
Notes
M1: Attempt correct process to find \(R\)
M1: Attempt correct process to find \(\tan\alpha\) (or equiv with \(\sin\alpha\) or \(\cos\alpha\))
Condone \(\cos\alpha = 3\), \(\sin\alpha = 4\) seen in method, leading to \(\tan\alpha = \frac{4}{3}\)
M0 for \(\tan\alpha = \frac{3}{4}\)
A1: Obtain \(5\cos(2x - 53.1)\). Could be implied by \(R = 5\) and \(\alpha = 53.1\) (awrt 53.1)
Condone not being written out in required form, and ISW any error when attempting to do so
A0 for \(\alpha = 0.927\) radians
| Scheme | Marks | AO |
|---|---|---|
| DR | ||
| \(5\cos(2x - 53.1) + 6 = 4(5\cos(2x - 53.1) - 1)\) \(15\cos(2x - 53.1) = 10\) | M1 | 3.1a |
| A1FT | 2.1 | |
| \(\cos(2x - 53.1) = \frac{2}{3}\) \(2x - 53.1 = 48.2\) \(2x = 101.3\) \(x = 50.7\) | M1* | 1.1 |
| A1 | 1.1 | |
| \(2x - 53.1 = -48.2,\ 48.2,\ 311.8,\ 408.2\) \(2x = 4.94,\ 101.3,\ 364.9,\ 461.3\) | M1dep* | 1.1 |
| \(x = 2.47,\ 50.7,\ 182.5,\ 230.7\) | A1 | 1.1 |
| A1 | 1.1 | |
| [7] |
Notes
M1: Clear fraction, attempt to rearrange and gather like terms
A1FT: Obtain correct harmonic equation, with like terms gathered. FT their \(R\cos(2x - \alpha)\)
Could still be in terms of \(\cos 2x\) and \(\sin 2x\) eg \(9\cos 2x + 12\sin 2x - 10\ (= 0)\)
M1*: Correct process to find at least one value for \(x\). Must be solving equation of the form \(\cos(2x \pm \alpha) = k\)
Allow substitution soi eg \(n\) = their \(5\cos(2x - 53.1)\) to obtain \(3n = 10\)
A1: Obtain at least one correct value. Likely to be using 48.2
Allow answers in range [50.6, 50.7] – see later A1 detail for range for other solutions
M1dep*: Correct process to find (at least) one further value for \(x\)
A1: Obtain one further correct value. Correct process to find \(x\) from a second value for their \(\cos^{-1}\left(\frac{2}{3}\right)\) Or their first solution \(+\ 180^\circ\) oe
Answers in range [2.46, 2.48], [182, 183], [230, 231]
A1: Obtain all 4 correct values, and no others in given range. Answers in range [2.46, 2.48], [182, 183], [230, 231]
Alt method if not harmonic form
| Scheme | Marks | AO |
|---|---|---|
| \(3\cos 2x + 4\sin 2x + 6 = 12\cos 2x + 16\sin 2x - 4\) \(9\cos 2x + 12\sin 2x - 10\ (= 0)\) | M1 | 3.1a |
| \(225\sin^2 2x - 240\sin 2x + 19\ (= 0)\) \(225\cos^2 2x - 180\cos 2x - 44\ (= 0)\) \(19\tan^2 x - 24\tan x + 1\ (= 0)\) \(900\cos^4 x - 1260\cos^2 x + 361\ (= 0)\) \(900\sin^4 x - 540\sin^2 x + 1\ (= 0)\) | A1FT | 2.1 |
| M1* | 1.1 | |
| A1 | 1.1 | |
| M1dep* | 1.1 | |
| \(x = 2.47,\ 50.7,\ 182.5,\ 230.7\) | A1 | 1.1 |
| A1 | 1.1 |
M1: Clear fraction, attempt to rearrange and gather like terms. Could have terms in \(\cos 2x\) and \(\sin 2x\) or even \(\tan x\), \(\cos x\), \(\sin x\), \(\cos x\sin x\) (if double angle formulae used)
A1FT: Obtain correct equation in a single trig ratio, with like terms gathered. Could be in terms of \(\sin x\) or \(\cos x\), if double angle formulae used but equation never seen in terms of just \(\cos 2x\) or \(\sin 2x\)
M1*: Correct process to find at least one value for \(x\). Solving a 3 term quadratic (or possibly quartic) in a single trig ratio (condone BC) and also attempt \(x\)
A1: Obtain at least one correct value. Allow answers in range [50.6, 50.7] – see later A1 detail for range for other solutions
M1dep*: Correct process to find one further value for \(x\). Must be finding a secondary value from their principal value. M0 if just finding principal value from other quadratic root
A1: Obtain one further correct value. Answers in range [2.46, 2.48], [182, 183], [230, 231]
A1: Obtain all 4 correct values, and no others in given range. Answers in range [2.46, 2.48], [182, 183], [230, 231]