# 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
Reputation: 0

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: 410
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 28

### 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
Reputation: 0

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
Reputation: 0

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

Math Tutor

Posts: 410
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 28

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

Math Tutor

Posts: 410
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 28

### Re: Please Solve this tricky Trignometric equation

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.

______________________________________________________________________________________________

AlfredJPK55

Posts: 1
Joined: Tue Mar 10, 2015 5:29 am
Reputation: 0

### Re: Please Solve this tricky Trignometric equation

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

leesajohnson

Posts: 208
Joined: Thu Dec 31, 2015 7:11 am
Location: London
Reputation: -33

### Re: Please Solve this tricky Trignometric equation

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.
Guest

### Re: Please Solve this tricky Trignometric equation

$$sin(x)+ tan(x)= \sqrt{3}$$
$$sin(x)+ \frac{sin(x)}{cos(x)}= \sqrt{3}$$
$$\frac{sin(x)(cos(x)+ 1)}{cos(x)}= \sqrt{3}$$
$$sin(x)= \frac{\sqrt{3}cos(x)}{cos(x)+ 1}$$
$$\sqrt{1- cos^2(x)}= \frac{\sqrt{3}cos(x)}{cos(x)+ 1}$$

Let y= cos(x)
$$\sqrt{1- y^2}= \frac{\sqrt{3}y}{y+ 1}$$

Square both sides
$$1- y^2= \frac{3y^2}{(y+ 1)^2}$$
$$(1-y^2)(y^2+2y+ 1)= 3y^2$$
$$2y+1-y^4-2y^3= 3y^2$$
$$y^4+ 2y^3+ 3y^2-2y- 1= 0$$

HallsofIvy

Posts: 143
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 57

Return to Trigonometry - sin, cos, tan, cot, arcsin, arccos, arctan, arccot

### Who is online

Users browsing this forum: No registered users and 1 guest