 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/2/b/solution

sin x = \frac{e ^{i x} - e ^{- i x}}{2 i}, cos x = \frac{e ^{i x} + e ^{- i x}}{2} .

Cubing the expression for the sine, and grouping the

e^{3 i x}

and

e^{- 3 i x}

terms, gives

sin^{3} x = - \frac{1}{4} (sin 3 x - 3 sin x), sin 3 x = 3 sin x - 4 sin^{3} x .

The required equation therefore becomes

sin x (1 - 4 sin^{2} x) = 0

. The complete real solution set is

x = nπ or x = nπ \pm \frac{π}{6}, n \in Z .

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