OCR MEI C2 2010 June — Question 8 5 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeQuadratic in sin²/cos²/tan²
DifficultyModerate -0.3 This is a standard trigonometric equation requiring the identity sin²θ + cos²θ = 1 to convert to a single trigonometric function, then solving a quadratic equation in cos²θ. The method is routine for C2 level with straightforward algebraic manipulation and finding angles in the given range, making it slightly easier than average but still requiring proper technique.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
Showing your method clearly, solve the equation \(4 \sin^2 \theta = 3 + \cos^2 \theta\), for values of \(\theta\) between \(0°\) and \(360°\). [5]
AnswerMarks Guidance
Subst. of \(1 - \cos^2 \theta\) or \(1 - \sin^2 \theta\)M1
\(5 \cos^2 \theta = 1\) or \(5 \sin^2 \theta = 4\)A1
\(\cos \theta = \pm\sqrt{\text{their }\frac{1}{5}}\) or \(\sin \theta = \pm\sqrt{\text{their }\frac{4}{5}}\) o.e.M1
\(63.4, 116.6, 243.4, 296.6\)B2 Accept to nearest degree or better; B1 for 2 correct (ignore any extra values in range) 
Subst. of $1 - \cos^2 \theta$ or $1 - \sin^2 \theta$ | M1 | 
$5 \cos^2 \theta = 1$ or $5 \sin^2 \theta = 4$ | A1 | 
$\cos \theta = \pm\sqrt{\text{their }\frac{1}{5}}$ or $\sin \theta = \pm\sqrt{\text{their }\frac{4}{5}}$ o.e. | M1 | 
$63.4, 116.6, 243.4, 296.6$ | B2 | Accept to nearest degree or better; B1 for 2 correct (ignore any extra values in range)
Showing your method clearly, solve the equation $4 \sin^2 \theta = 3 + \cos^2 \theta$, for values of $\theta$ between $0°$ and $360°$. [5]

\hfill \mbox{\textit{OCR MEI C2 2010 Q8 [5]}}
This paper (12 questions)
View full paper
Q1 2 Q2 4 Q3 5 Q4 4 Q5 4 Q6 5 Q7 2 Q8 5 Q9 5 Q10 13 Q11 13 Q12 10