| Exam Board | CAIE |
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| Module | P3 (Pure Mathematics 3) |
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| Year | 2013 |
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| Session | November |
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| Topic | Harmonic Form |
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7
- Given that \(\sec \theta + 2 \operatorname { cosec } \theta = 3 \operatorname { cosec } 2 \theta\), show that \(2 \sin \theta + 4 \cos \theta = 3\).
- Express \(2 \sin \theta + 4 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\) where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) correct to 2 decimal places.
- Hence solve the equation \(\sec \theta + 2 \operatorname { cosec } \theta = 3 \operatorname { cosec } 2 \theta\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).