Secant issues

Solve 2Sec^2=5tanx-1
I assume you meant

Solve 2(Sec x)^2=5tanx-1

It's got me stumped too for the moment. But I am still playing with it.

Solve 2(Sec x)^2=5tanx-1

I tried the usual method and it didn't work for me so I decided to get a bit inventive.

I said, let t=tanx
now you can draw a right angled triangle and let x be one of the acute angles.
let 't' be the opposite side and 1 be the adjacent side.
Now you have a triangle where tanx = t
Use pythagoras to get the hypotenuse = Sqr(1+t^2)

So now sec x = [ Sqr(1+t^2) ] / 1

2sec^2 x = 2(1+t^2)
So
2(1+t^2) = 5t - 1
rearrange and factorise and you get
(2t-3)(t-1) = 0
so t = 3/2 or t = 1
t=tanx

If t=1 then x = pi/4 +- n*pi Where n is an integer

If t=3/2 then x = atan(3/2) +- n*pi

Hence
x = pi/4 +- n*pi OR x = atan(3/2) +- n*pi Where n is an integer (atan stands for arc tan which is the same as inverse tan)

I did consider this a little more, sometimes 1 or more of answers gained like this can be wrong because of negative signs (you do need to be careful) but
2(Sec x)^2 +1 is always going to be positive so x needs to be in the 1st or 3rd quadrant so I don't think there is any need to be concerned.

It had me stumped too! Interesting answer although.

You may of just confirmed my suspicions that our teacher may of made an error with this question, it isn't something we've quite come across, not at this difficulty anyway.

Hi Ant Nix125,

Maybe your teacher found an easier method.
But yes, it may have been set in error.
it was pretty difficult.

(ofcourse my answer may be wrong but barring stupid errors I think it is sound.)

I am glad you sent it my way. If there is more feed back from your teacher I would like you to continue the post.