cosec x = – 2 ⇒ sin x = −

1

2

A value of x, satisfying sinx =

1

2

is

π

6

Principal Solution :We have, sinx = −

1

2

Thus, sin (π+

π

6

) = - sin

π

6

=

−1

2

and , sin (2π−

π

6

) = - sin

π

6

= −

1

2

Thus, sin

7π

6

= sin (

11π

6

) = −

1

2

Therefore principal solutions are

7π

6

,

11π

6

General Solution : sinx = −

1

2

⇒ sin x = sin (π+

π

6

)
⇒ sin x = sin

7π

6

⇒ x = nπ + (−1)n

7π

6

, n ∊ Z
Since if sin x = sin α , then x = nπ + (−1)n α , n ∊ Z