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

Expert Answers

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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` .

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