Question

Solve the equation `tan2x+1=sec2x`?

Answer 1

I am sorry. It is incorrect. The correct answer is `x = npi, n in ZZ`

Explanation:

Given: `tan(2x)+1=sec(2x)`

You can write this as:

`tan(2x)+1-sec(2x)=0`

This is the graph below:

graph{tan(2x)+1-sec(2x) [-10, 10, -5, 5]}

Please observe that the graph crosses zero at every integer multiple of `pi`.

This makes the answer `x = npi` where an n is any member of the set of integer numbers.

Written a bit more formally: `x = npi, x in ZZ`

Answer 2

Please see below.

Explanation:

You have been correct but

let me list out steps taken by you and what has gone wrong.

`tan2x+1 =(sin2x)/(cos2x) +1` and `sec2x=1/(cos2x)`

Therefore `(sin2x+cos2x)/(cos2x) = 1/(cos2x)`

or `sin2x+cos2x=1`

or `sin2x+cos2x-1=0`

or `sin2x- 2sin^2x=0`

or `2sinxcosx-2sin^2x=0`

or `2sinx(cosx-sinx)=0`

Therefore either `2sinx=0` i.e. `sinx=0` and `x=npi`

or `cosx-sinx=0` i.e. `tanx=1` and `x=npi+pi/4`

But for `tan(2xxpi/4)` and `sec(2xxpi/4)`, as `2xxpi/4=pi/2` and for this these ratios are not defined

Solution within `0 <= x < 2pi` is `x={0,pi,2pi}`

or general solution is `x=npi`.