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Feroking · Jan 17, 2011, 02:34 PM

I need to find all solution between [0,2pi]

ebaines · Jan 17, 2011, 02:38 PM

Please do not double post questions. You asked before, and it was answered here: https://www.askmehelpdesk.com/mathem...2x-544916.html

Feroking · Jan 17, 2011, 03:03 PM

You didn't even answer it lol you just stated basic steps I already knew jeez

ebaines · Jan 17, 2011, 03:16 PM

Well asking the same question a 2nd time doesn't help. If you already knew those steps, then be more specific - show us how far you got in solving it and where you ran into difficulty. Then you can get a more specific guidance.

Feroking · Jan 17, 2011, 03:21 PM

OK I got to the point in which its 1+(1/cos^2(x))sin^2(x)=2x then I Change it to 1+(1/cos^2(x))(1-cos^2(x)=2x and now I'm stuck...

ebaines · Jan 17, 2011, 03:33 PM

OK. First, since sinx/cosx = tan x, you have

1+ tan^2x = 2x

This does NOT factor out as (1+tanx)(1-tanx) = 2x. What you can do is this:

$ 1 + \tan^2x = 2x \\ \\ \frac {\cos^2x + \sin^2x} {\cos^2x} = 2x \\ \\ \frac 1 {cos^2x} = 2x \\ 2x \cos^2x = 1 $

Hmm.. not much more you can do here to get a closed form solution. Are you familiar with any approximation techniques to estaime the value for x?

Alternatively, I'm just wondering if the problem shouldn't have been written with a minus sign rather than the plus sign, like this:

$ 1 - \sec^2(x) \sin^2x=2x $

This has a much easier solution.