"The Humongus Book of Trigonometry Problems" (by Kelley) has the problem

sin (x - pi/3) = 0

as one of their "Advanced Trigonometric Equations". After a bunch of steps, the answers for between 0 and 2*pi as:

pi/3, 2pi/3, 4pi/3, 5pi/3

How can 2pi/3 be a solution? I plug in and get sin(pi/3) and that is not zero. A similar problem with 5pi/2.