Question 5 - A-Level Maths

Edexcel C3 — Question 5 9 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Standard trigonometric equations
Type	Convert to quadratic in tan
Difficulty	Challenging +1.2 This question requires applying the tangent addition formula, algebraic manipulation to form a quadratic in tan x, and solving within a specified interval. While it involves multiple steps and careful algebraic work, the techniques are standard C3 content with no novel insight required. The 9-mark allocation reflects computational length rather than conceptual difficulty, placing it moderately above average.
Spec	1.05l Double angle formulae: and compound angle formulae 1.05o Trigonometric equations: solve in given intervals

Find the values of $x$ in the interval $-180 < x < 180$ for which $$\tan (x + 45)^{\circ} - \tan x^{\circ} = 4,$$ giving your answers to 1 decimal place. [9]

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Answer	Marks	Guidance
$\frac{\tan x + \tan 45°}{1-\tan x \tan 45°} = \tan x = 4$	M1
$\frac{\tan x + 1}{1 - \tan x} = 4 + \tan x$	A1
$\tan x + 1 = (4 + \tan x)(1 - \tan x)$	A1
$\tan^2 x + 4 \tan x - 3 = 0$	A1
$\tan x = \frac{-4 \pm \sqrt{16+12}}{2} = -2 \pm \sqrt{7}$	M1
$x = 180 - 77.9, -77.9$ or $32.9, -180 + 32.9$	B1 M1
$x = -147.1, -77.9, 32.9, 102.1$ (1dp)	A2	(9 marks)

$\frac{\tan x + \tan 45°}{1-\tan x \tan 45°} = \tan x = 4$ | M1 |
$\frac{\tan x + 1}{1 - \tan x} = 4 + \tan x$ | A1 |
$\tan x + 1 = (4 + \tan x)(1 - \tan x)$ | A1 |
$\tan^2 x + 4 \tan x - 3 = 0$ | A1 |
$\tan x = \frac{-4 \pm \sqrt{16+12}}{2} = -2 \pm \sqrt{7}$ | M1 |
$x = 180 - 77.9, -77.9$ or $32.9, -180 + 32.9$ | B1 M1 |
$x = -147.1, -77.9, 32.9, 102.1$ (1dp) | A2 | (9 marks)

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Find the values of $x$ in the interval $-180 < x < 180$ for which
$$\tan (x + 45)^{\circ} - \tan x^{\circ} = 4,$$
giving your answers to 1 decimal place. [9]

\hfill \mbox{\textit{Edexcel C3  Q5 [9]}}

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Q1 6 Q2 7 Q3 8 Q4 9 Q5 9 Q6 10 Q7 12 Q8 14

Answer	Marks	Guidance
\(\frac{\tan x + \tan 45°}{1-\tan x \tan 45°} = \tan x = 4\)	M1
\(\frac{\tan x + 1}{1 - \tan x} = 4 + \tan x\)	A1
\(\tan x + 1 = (4 + \tan x)(1 - \tan x)\)	A1
\(\tan^2 x + 4 \tan x - 3 = 0\)	A1
\(\tan x = \frac{-4 \pm \sqrt{16+12}}{2} = -2 \pm \sqrt{7}\)	M1
\(x = 180 - 77.9, -77.9\) or \(32.9, -180 + 32.9\)	B1 M1
\(x = -147.1, -77.9, 32.9, 102.1\) (1dp)	A2	(9 marks)