You need to group the terms such that:
`(2secxtanx+2secx)+(tan x + 1) = 0 `
You need to factor out `2sec x` in the first group such that:
`2sec x(tan x +1) + (tan x + 1) = 0 `
You need to factor out `tan x + 1` such that:
`(tan x + 1)(2sec x + 1) = 0`
You need to consider the first factor equal to zero such that:
`tan x +1 = 0 => tan x = -1 => x = arctan(-1) + npi`
Using the trigonometric identity `arctan(-alpha) = -arctan alpha` yields:
`x = -arctan 1 + npi => x = -pi/4 + npi`
You need to consider the second factor equal to zero such that:
`2sec x + 1 = 0 => 2sec x = -1 => sec x = -1/2`
Since `sec x = 1/cos x => 1/cos x = -1/2 => cos x = -2`
The equation `cos x = -2` is invalid since the values of cos x cannot be smaller than -1.
Hence, evaluating the general solutions to the given trigonometric equation yields `x = -pi/4 + npi` .