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Aus Online Mathematik Brückenkurs 1

If we use the trigonometric relation sin(−x)=−sinx, the equation can be rewritten as

sin2x=sin(−x).

In exercise 4.4:5a, we saw that an equality of the type

sinu=sinv

is satisfied if

u=v+2noru=−v+2n

where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy

2x=−x+2nor2x=−(−x)+2n

i.e.

3x=2norx=+2n.

The solutions to the equation are thus

xx=32n=+2n

where n is an arbitrary integer.