# Math Help - Trigonometric Equation

1. ## Trigonometric Equation

Hi everyone,

I've been given the following equation:

$3cot(2x)+tan(x)-4+\frac{3}{sin^2(x)}=0$
and been asked to solve it the interval
$[0,\frac{3\pi }{2}[$
I've spent hours basically going in circles, unable to solve it. If anyone could show me how to simplfy this I would really appreciate it.

Thanks

2. ## Re: Trigonometric Equation

My strategy would be to reduce to a single trig function. First, $cot(a)=\frac{cos(a)}{sin(a)}$ so that $cos(2x)= \frac{cos(2x)}{sin(2x)}= \frac{cos^2(x)- sin^2(x)}{2sin(x)cos(x)}= \frac{1- 2sin^2(x)}{2sin(x)\sqrt{1- sin^2(x)}}$. Similarly, $tan(x)= \frac{sin(x)}{cos(x)}= \frac{sin(x)}{\sqrt{1- sin^2(x)}}$. So this equation says
$\frac{3(1- 2sin^2(x))}{2sin(x)\sqrt{1- sin^2(x)}}+ \frac{sin(x)}{\sqrt{1- sin^2(x)}}- 4+ \frac{3}{sin^2(x)}= 0$

"Clear the fractions" by multiplying through by $2sin^2(x)\sqrt{1- sin^2(x)}$:
$3sin(x)(1- 2sin^2(x)+ 2sin^3(x)- 8sin^2(x)\sqrt{1- sin^2(x)}+ 6\sqrt{1- sin^2(x)}= 0$

Let y= sin(x) so this becomes $3y(1- 2y^2)+2y^3- 8y^2\sqrt{1- y^2}+ 6\sqrt{1- y^2}= 0$

We can write this as $3y(1- y^2)+ 2y^3= (8y^2+ 6)\sqrt{1- y^2}$
and squaring both sides gives a sixth degree polynomial in y.

3. ## Re: Trigonometric Equation

Thanks a lot for your help! It's much appreciated.

I think you may have made two small errors (either that or I misunderstood something)

I think (8y2+6)√(1-y2)
should rightly be (8y2-6)√(1-y2)

and you also dropped the 2 from 3y(1-2y3) at the end. Unless I made some mistake.

In any case doing things the way you showed me I ended up with

16sin6x-88sin4x+105sin2x-36

which I then factored and got 3 facts, 2 of which were the same, and one of which was impossible (sin(x)=±2 when simplified)
In any case the final answer, if I didn't mess up somewhere along the way, is x=pi/3, 2pi/3, or 4pi/3 in the given interval.

Thanks a lot for your help, I don't know if I ever would have found the answer without you.
Oh and sorry, for some reason I am heaving trouble with latex all of a sudden. I'll look into that later though.

4. ## Re: Trigonometric Equation

Could you check that you've copied the equation correctly ?
If that second term were $2\tan x$ I think it drops out quite nicely.

5. ## Re: Trigonometric Equation

I did indeed copy it correctly. The question from a practice exam for my universities entrance exam and I think it not dropping out nicely is expressly their purpose unfortunately.

6. ## Re: Trigonometric Equation

Originally Posted by DCB
Thanks a lot for your help! It's much appreciated.

I think you may have made two small errors (either that or I misunderstood something)

I think (8y2+6)√(1-y2)
should rightly be (8y2-6)√(1-y2)

and you also dropped the 2 from 3y(1-2y3) at the end. Unless I made some mistake.

In any case doing things the way you showed me I ended up with

16sin6x-88sin4x+105sin2x-36

which I then factored and got 3 facts, 2 of which were the same, and one of which was impossible (sin(x)=±2 when simplified)
In any case the final answer, if I didn't mess up somewhere along the way, is x=pi/3, 2pi/3, or 4pi/3 in the given interval.

Thanks a lot for your help, I don't know if I ever would have found the answer without you.
Oh and sorry, for some reason I am heaving trouble with latex all of a sudden. I'll look into that later though.
Just for information Mathematica is showing an additional solution in this interval at $x=2\arctan\left(\frac{1}{2}\left(1+\sqrt{5}\right) \right)$

7. ## Re: Trigonometric Equation

Really? Any idea how I could go abut finding that answer? And also unless I am mistaken isn't that the golden ratio?

9. ## Re: Trigonometric Equation

It certainly is. Nice catch.

Interesting. Ibdutt's cubic is also showing 4 solutions including $x=\pi-\arctan(2)$ which turns out to be equal to the golden ratio solution but in a nicer form.