# How do I solve 2secxtanx+2secx+tanx+1=0

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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` .**

2secx tanx + 2secx + tanx + 1 = 0

tanx(2secx + 1) + 2secx+1 = 0

(2secx + 1)(tanx + 1) = 0

2secx + 1 = 0 or tanx + 1 = 0

secx = -1/2 or tanx = -1

tanx = -1 (ignoring secx = -1/2 which is not a valid value)

hope this will help you.../ just do pratice ... ... ... ... .. (: