Solve 2sin^2x=sinx, if 0

sin

x

(

2

sin

x

+

1

)

=

0

Use trig table and unit circle - ->

a. sin x = 0 = > x = 0, and

x

=

π

and

x

=

2

π

b. 2sin x + 1 = 0 = >

sin

x

=

-

1

2

=>

x

=

7

π

6

and

x

=

11

π

6

.

Finally, within interval

[

0

,

2

π

]

,

there are 5 answers:

0

,

π

,

2

π

,

7

π

6

and

11

π

6