Edexcel C2 — Question 6 8 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeConvert to quadratic in sin/cos
DifficultyStandard +0.3 This is a trigonometric equation requiring the Pythagorean identity (sin²θ + cos²θ = 1) to convert to a quadratic in cosθ, then solving the quadratic and finding angles in the full range. It's slightly above average difficulty due to the algebraic manipulation and multiple solutions, but follows a standard C2 pattern with no novel insight required.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
Find, in degrees, the value of \(\theta\) in the interval \(0 \leq \theta < 360°\) for which $$2\cos^2\theta - \cos\theta - 1 = \sin^2\theta.$$ Give your answers to \(1\) decimal place where appropriate. [8]
AnswerMarks Guidance
\(2\cos^2\theta - \cos\theta - 1 = 1 - \cos^2\theta\)M1
\(3\cos^2\theta - \cos\theta - 2 = 0\)A1
\((3\cos\theta + 2)(\cos\theta - 1) = 0\)M1 A1 \(\cos\theta = -\frac{2}{3}\) or \(1\)
\(\theta = 0\)B1 A1 \(\theta = 131.8°\)
\(\theta = (360 - "131.8°") = 228.2°\)M1 A1 ft 
$2\cos^2\theta - \cos\theta - 1 = 1 - \cos^2\theta$ | M1 |

$3\cos^2\theta - \cos\theta - 2 = 0$ | A1 |

$(3\cos\theta + 2)(\cos\theta - 1) = 0$ | M1 A1 | $\cos\theta = -\frac{2}{3}$ or $1$

$\theta = 0$ | B1 A1 | $\theta = 131.8°$

$\theta = (360 - "131.8°") = 228.2°$ | M1 A1 ft |
Find, in degrees, the value of $\theta$ in the interval $0 \leq \theta < 360°$ for which
$$2\cos^2\theta - \cos\theta - 1 = \sin^2\theta.$$

Give your answers to $1$ decimal place where appropriate. [8]

\hfill \mbox{\textit{Edexcel C2  Q6 [8]}}
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