sec x = 2 ⇒ cos x =

1

2

A value of x satisfying cosx = /2 is

π

3

Principal Solution:
We know that, cos

π

3

=

1

2

and cos (2π−

π

3

) = cos

π

3

=

1

2

⇒ Thus, cos

π

3

= cos

5π

3

=

1

2

Therefore principal solutions are

π

3

,

6π

3

General Solution : We have,
cos x =

1

2

⇒ cos x = cos

π

3

⇒ x = 2nπ ±

π

3

, n ∊ Z
Since If cos θ = cos α , then θ = 2nπ ± α , n ∊ Z