Please Solve this tricky Trignometric equation

[parmar.abhishek]

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

[Math Tutor]

Maybe there are many ways to solve this trigonometric equation:

I use the following trigonometric formulas:

[tex]sin2u = \frac{2tan^2u}{1+tan^2u}[/tex]
and
[tex]tan2u = \frac{2tan^2u}{1 - tan^2u}[/tex]

so:

[tex]\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}[/tex]

[tex]\frac{4tan^2\frac{x}{2} }{1-tan^4\frac{x}{2}} = \sqrt{3}[/tex]


now replace [tex]tan^2\frac{x}{2} = t[/tex]
and solve the quadratic equation.

[parmar.abhishek]

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.

[parmar.abhishek]

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.

[Math Tutor]

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

[Math Tutor]

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.

[AlfredJPK55]

When I saw the arrangement, I was thinking myself as a nitwit not ready to do such simple steps and was cheerful to see the arrangement.

At the same time when I found that the recipe utilized isn't right, I am again remain faithful to this issue.

Also I am holding up for somebody who can help me in any capacity.






______________________________________________________________________________________________
NoorAlamShahzad

[leesajohnson]

Sin2u = 2tan²u/1+tan²u
and
tan2u = 2tan²u/1-tan²u

so we will put sin2u and tan2u values into the given equation Sin x + tan x = √3

after solving this equation we will get

4tan²x/2 / 1-tan4 x/2 = √3
now we will replace tan²x/2 by t

4t / 1-t² = √3

[Guest]

Its not a wrong question! it is right question and if you want to check the solution of that then wait i will soon upload that.

[HallsofIvy]

[tex]sin(x)+ tan(x)= \sqrt{3}[/tex]
[tex]sin(x)+ \frac{sin(x)}{cos(x)}= \sqrt{3}[/tex]
[tex]\frac{sin(x)(cos(x)+ 1)}{cos(x)}= \sqrt{3}[/tex]
[tex]sin(x)= \frac{\sqrt{3}cos(x)}{cos(x)+ 1}[/tex]
[tex]\sqrt{1- cos^2(x)}= \frac{\sqrt{3}cos(x)}{cos(x)+ 1}[/tex]

Let y= cos(x)
[tex]\sqrt{1- y^2}= \frac{\sqrt{3}y}{y+ 1}[/tex]

Square both sides
[tex]1- y^2= \frac{3y^2}{(y+ 1)^2}[/tex]
[tex](1-y^2)(y^2+2y+ 1)= 3y^2[/tex]
[tex]2y+1-y^4-2y^3= 3y^2[/tex]
[tex]y^4+ 2y^3+ 3y^2-2y- 1= 0[/tex]