# Solve the following trigonometric equation: tan x + sec 2x = 1 ; [-pi/2 ≤ x ≤ pi/2]

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The following trigonometric equation: tan x + sec 2x = 1 has to be solved for x with the following constraints [-pi/2 ≤ x ≤ pi/2].

tan x + sec 2x = 1

=> `tan x + 1/(cos 2x) = 1`

=> `tan x*cos 2x + 1 = cos 2x`

=> `tan x*(2*cos^2x - 1) + 1 = cos 2x`

=> `(sin x/cosx)*(2*cos^2x - 1) = 1 - 2*sin ^2x - 1`

=> `(sin x/cos x)(2*cos^2x - 1) = -2*sin^2x`

=> `(2*cos^2x)/cosx - 1/cos x = -2*sin x`

=> `2*cos x - 1/cos x = -2*sin x`

=> `2*cos^2x - 1 = -2*sin x*cos x`

=> `cos 2x = -sin 2x`

=> `tan 2x = -1`

2x = `tan^-1(-1)`

=> 2x = -45 degrees

=> x = -22.5 degrees

**The solution of the equation is x = -22.5 degrees.**

How did you get 0? :(

0 and 3pi/8 are also answers.