 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/4/solution

In one period, sine equals

- 3 /2

4 π /3

and

5 π /3

, and lies below that value between them. Since sine is a periodic function with period

2 π

, all solutions are:

x \in k \in Z ⋃ [\frac{4 π}{3} + 2 πk, \frac{5 π}{3} + 2 πk] .

The endpoints are included because the inequality is non-strict.

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