| Exam Board | OCR |
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| Module | C2 (Core Mathematics 2) |
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| Year | 2010 |
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| Session | January |
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| Topic | Trig Equations |
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1
- Show that the equation
$$2 \sin ^ { 2 } x = 5 \cos x - 1$$
can be expressed in the form
$$2 \cos ^ { 2 } x + 5 \cos x - 3 = 0$$
- Hence solve the equation
$$2 \sin ^ { 2 } x = 5 \cos x - 1$$
giving all values of \(x\) between \(0 ^ { \circ }\) and \(360 ^ { \circ }\).