Sets of solutions of trigonometic equations

Question

Here is a part of my textbook.Here sets A and B are shown equal.But $\frac{2π}{3}$ does not belongs to Set A but belongs to Set B.Then how does they both are equal?

Answer 1

There is a single typographic error--a stray comma--in the statement in the book.

What is written in the book is

$$B = \left\{ \ldots, \pm \frac\pi6, \frac{2\pi}3, \pm \frac\pi6, \ldots \right\}.$$

What should have been written is

$$B = \left\{ \ldots, \pm \frac\pi6, \frac{2\pi}3 \pm \frac\pi6, \ldots \right\}.$$

Notice that in the book there is a comma with $\frac{2\pi}3$ on the left side of the comma and $\pm \frac\pi6$ on the right side. That comma should not be there and in the corrected formula I have removed it. Thus $\frac{2\pi}3$ is not a member of $B$; instead, $\frac{2\pi}3 - \frac\pi6$ is a member of $B$ (by taking the $-$ sign for the $\pm$ in $\frac{2\pi}3 \pm \frac\pi6$) and $\frac{2\pi}3 + \frac\pi6$ is a member of $B$ (by taking the $+$ sign for the $\pm$).

Notice that $\frac{2\pi}3 - \frac\pi6 = \frac\pi2$ (which is listed as an element of $A$) and $\frac{2\pi}3 + \frac\pi6 = \frac{5\pi}6$ (which also is listed as an element of $A$).

It is good to check these things to make sure they are really true! Indeed in this case you found an error. This can happen as a book is prepared for printing.