Please Solve this tricky Trignometric equation

by **parmar.abhishek** » Sun Sep 19, 2010 5:40 am

I have tried in many ways to solve this equation.
Please help me on the solution. This problem is from R.D Sharma

Sin x + tan x = √3

by **Math Tutor** » Sun Sep 19, 2010 9:33 am

Maybe there are many ways to solve this trigonometric equation:

I use the following trigonometric formulas:

$sin2u = \frac{2tan^2u}{1+tan^2u}$
and
$tan2u = \frac{2tan^2u}{1 - tan^2u}$

so:

$\frac{2tan^2\frac{x}{2} }{1+tan^2\frac{x}{2}} + \frac{2tan^2\frac{x}{2} }{1-tan^2\frac{x}{2}} = \sqrt{3}$

$\frac{4tan^2\frac{x}{2} }{1-tan^4\frac{x}{2}} = \sqrt{3}$

now replace $tan^2\frac{x}{2} = t$
and solve the quadratic equation.

by **parmar.abhishek** » Mon Sep 20, 2010 12:54 pm

the formula used for tan 2u is not correct.

Please check that again.
I am tired of this solution please someone help me with this problem.

by **parmar.abhishek** » Wed Sep 22, 2010 1:27 pm

When I saw the solution, I was thinking myself as a fool not able to do such easy steps and was happy to see the solution.
But when I discovered that the formula used is wrong, I am again stuck with this problem.
And I am waiting for someone who can help me in any way.

by **Math Tutor** » Fri Sep 24, 2010 6:00 am

The both formulas are wrong. it is 2tan not 2tan²
I will think once again.

by **Math Tutor** » Wed Sep 29, 2010 1:29 am

Here is the discussion in the bulgarian math forum:
http://www.math10.com/f/viewtopic.php?f=49&t=3198

You could use google translate.

Please Solve this tricky Trignometric equation

Please Solve this tricky Trignometric equation

wrong formula used

Please anyone reply

Who is online