Edexcel C2 — Question 4 8 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeConvert to quadratic in sin/cos
DifficultyStandard +0.8 This question requires multiple steps: using the Pythagorean identity to convert sin²x to (1-cos²x), rearranging to form a quadratic in cos x, solving the quadratic, then finding all solutions in the given interval including consideration of the CAST diagram. While the individual techniques are standard C2 content, the initial algebraic manipulation and need to find multiple solutions elevates this above a routine trigonometric equation.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
Find all values of \(x\) in the interval \(0 \leq x < 360°\) for which $$2\sin^2 x - 2\cos x - \cos^2 x = 1.$$ [8]
AnswerMarks Guidance
\(2(1 - \cos^2 x) - 2\cos x - \cos^2 x = 1\)M1
\(3\cos^2 x + 2\cos x - 1 = 0\)A1
\((3\cos x - 1)(\cos x + 1) = 0\)M1
\(\cos x = -1\) or \(\frac{1}{3}\)A1
\(x = 180°\) or \(70.5°, 360° - 70.5°\)B2 M1
\(x = 70.5° (1\text{dp}), 180°, 289.5° (1\text{dp})\)A1 (8) 
$2(1 - \cos^2 x) - 2\cos x - \cos^2 x = 1$ | M1 |
$3\cos^2 x + 2\cos x - 1 = 0$ | A1 |
$(3\cos x - 1)(\cos x + 1) = 0$ | M1 |
$\cos x = -1$ or $\frac{1}{3}$ | A1 |
$x = 180°$ or $70.5°, 360° - 70.5°$ | B2 M1 |
$x = 70.5° (1\text{dp}), 180°, 289.5° (1\text{dp})$ | A1 | (8)
Find all values of $x$ in the interval $0 \leq x < 360°$ for which
$$2\sin^2 x - 2\cos x - \cos^2 x = 1.$$ [8]

\hfill \mbox{\textit{Edexcel C2  Q4 [8]}}
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