# Please Solve this tricky Trignometric equation

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

### Please Solve this tricky Trignometric equation

I have tried in many ways to solve this equation.

Sin x + tan x = √3

parmar.abhishek

Posts: 3
Joined: Sun Sep 19, 2010 5:37 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 $tan^2\frac{x}{2} = t$
Math Tutor

Posts: 312
Joined: Sun Oct 09, 2005 11:37 am

### wrong formula used

the formula used for tan 2u is not correct.

I am tired of this solution please someone help me with this problem.
parmar.abhishek

Posts: 3
Joined: Sun Sep 19, 2010 5:37 am

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.
parmar.abhishek

Posts: 3
Joined: Sun Sep 19, 2010 5:37 am

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

Posts: 312
Joined: Sun Oct 09, 2005 11:37 am

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

Math Tutor