sin(2x) = sinx
=2sinxcosx – sinx = 0
=sinx(2cosx – 1) = 0
=sinx = 0 or cosx = 1/2
If sinx = 0, then x = kπ, k = 0, ±1, ±2,…
Note :
If cosx = 1/2, then x = π/3 + 2kπ, 5π/3 + 2kπ, k = 0, ±1, ±2, …
The solutions in the interval [0, 2π] are:
0, π, 2π, π/3, 5π/3
