Find the general solutions of the following equation: `sqrt(3)sec2theta=2`

Explanation

Given,`sqrt(3)sec2theta=2`

`1cos2theta=sqrt(3)​/2⇒cos2theta=cos(π/6)`


using formula if
`cosx=cosy `then `x=2nπ±y` where `n=0,±1,±2......`

We get
`2theta=2npi±pi/6` where `n=0,±1,±2.......`

`theta=nπ±π/12` where `n=0,±1,±2,........​`