cot x = − √3 ⇒ tan x = −

1

√3

A value of x satisfying tanx =

1

√3

is

π

6

Principal Solution : We have tanx = −

1

√3

Thus , tan (π−

π

6

) = - tan

π

6

=

−1

√3

and , tan (2π−

π

6

) = - tan

π

6

= -

1

√3

Thus , tan

5π

6

= tan

11π

6

=

−1

√3

Therefore principal solutions are

5π

6

,

11π

6

General Solution :
We have, tanx =

−1

√3

⇒ tan x = tan (π−

π

6

) = tan

5π

6

⇒ x = nπ +

5π

6

, n ∊ Z
Since tan x = tan α , then x = nπ + α , n ∊ Z