Roman C. answered 02/27/17
Tutor
5.0
(845)
Masters of Education Graduate with Mathematics Expertise
We have, sin 6x= -sin x.
Thus either 6x = 2kπ - x or 6x = (2k+1)π + x for some integer k.
We can set different integers for k in both cases and get all solutions in [0,2π).
They are:
0
π/5
2π/7
4π/7
3π/5
6π/7
π
8π/7
7π/5
10π/7
12π/7
9π/5
2π/7
4π/7
3π/5
6π/7
π
8π/7
7π/5
10π/7
12π/7
9π/5