OCR MEI C2 2009 June — Question 7 5 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2009
SessionJune
Marks5
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TopicQuadratic trigonometric equations
TypeShow then solve: sin²/cos² substitution
DifficultyModerate -0.8 This is a straightforward C2 trigonometric equation requiring only the Pythagorean identity cos²θ = 1 - sin²θ to convert to quadratic form, then factorising sin θ(4sin θ - 1) = 0 and finding solutions in a restricted range. The 'show that' part is routine algebraic manipulation with no problem-solving required, making this easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
7 Show that the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) may be written in the form $$4 \sin ^ { 2 } \theta - \sin \theta = 0$$ Hence solve the equation \(4 \cos ^ { 2 } \theta = 4 - \sin \theta\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\). 
7 Show that the equation $4 \cos ^ { 2 } \theta = 4 - \sin \theta$ may be written in the form

$$4 \sin ^ { 2 } \theta - \sin \theta = 0$$

Hence solve the equation $4 \cos ^ { 2 } \theta = 4 - \sin \theta$ for $0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }$.

\hfill \mbox{\textit{OCR MEI C2 2009 Q7 [5]}}
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