Solution 4.4:6c

From Förberedande kurs i matematik 1

If we use the trigonometric relation sin −x=−sin x , 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+2n or u=−v+2n

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

2x=−x+2n or 2x=−−x+2n 

i.e.

3x=2n or x=+2n

The solutions to the equation are thus

x=32nx=+2n  ( n an arbitrary integer)