Hi everyone,
I've been given the following equation:
and been asked to solve it the interval
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
Hi everyone,
I've been given the following equation:
and been asked to solve it the interval
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
My strategy would be to reduce to a single trig function. First, $\displaystyle cot(a)=\frac{cos(a)}{sin(a)}$ so that $\displaystyle cos(2x)= \frac{cos(2x)}{sin(2x)}= \frac{cos^2(x)- sin^2(x)}{2sin(x)cos(x)}= \frac{1- 2sin^2(x)}{2sin(x)\sqrt{1- sin^2(x)}}$. Similarly, $\displaystyle tan(x)= \frac{sin(x)}{cos(x)}= \frac{sin(x)}{\sqrt{1- sin^2(x)}}$. So this equation says
$\displaystyle \frac{3(1- 2sin^2(x))}{2sin(x)\sqrt{1- sin^2(x)}}+ \frac{sin(x)}{\sqrt{1- sin^2(x)}}- 4+ \frac{3}{sin^2(x)}= 0$
"Clear the fractions" by multiplying through by $\displaystyle 2sin^2(x)\sqrt{1- sin^2(x)}$:
$\displaystyle 3sin(x)(1- 2sin^2(x)+ 2sin^3(x)- 8sin^2(x)\sqrt{1- sin^2(x)}+ 6\sqrt{1- sin^2(x)}= 0$
Let y= sin(x) so this becomes $\displaystyle 3y(1- 2y^2)+2y^3- 8y^2\sqrt{1- y^2}+ 6\sqrt{1- y^2}= 0$
We can write this as $\displaystyle 3y(1- y^2)+ 2y^3= (8y^2+ 6)\sqrt{1- y^2}$
and squaring both sides gives a sixth degree polynomial in y.
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 (8y^{2}+6)√(1-y^{2})
should rightly be (8y^{2}-6)√(1-y^{2})
and you also dropped the 2 from 3y(1-2y^{3}) at the end. Unless I made some mistake.
In any case doing things the way you showed me I ended up with
16sin^{6}x-88sin^{4}x+105sin^{2}x-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.