# Thread: Rearranging Equations and Trigonometry

1. ## Rearranging Equations and Trigonometry

I am stuck on these last three:

Question #1

$
\frac{tan^2+1}{sec(x)csc(x)}
$

-tan^2(x)+1= sec^2(x).
-sec/csc= tan (x)

Question#2

$3cot^2(x)=1$

$
0\leq x \leq 2 \pi
$

Solve the following for x

-Divide both sides by 3
- take the sq. root
-cot(x)=sqrt3
-tan of sqrt3 = pi/3

Question#3.

Rearrange the following equation to give an expression for dydx in terms of x and y.

$
1-3xy^2\frac{dy}{dx}-6x^2+\frac{dy}{dx}=5xy\frac{dy}{dx}-3x+6x^2\frac{dy}{dx}
$

Lost! Where do I begin?!

TIA as always!

2. Originally Posted by mvho
I am stuck on these last three:

Question #1

$
\frac{tan^2+1}{sec(x)csc(x)}
$

-tan^2(x)+1= sec^2(x).
-sec/csc= tan (x) ... ok

Question#2

$3cot^2(x)=1$

$
0\leq x \leq 2 \pi
$

Solve the following for x

-Divide both sides by 3
- take the sq. root
-cot(x)=sqrt3
-tan of sqrt3 = pi/3

$\textcolor{red}{\cot{x} = \pm \frac{1}{\sqrt{3}}}$

4 solutions.

Question#3.

Rearrange the following equation to give an expression for dydx in terms of x and y.

$
1-3xy^2\frac{dy}{dx}-6x^2+\frac{dy}{dx}=5xy\frac{dy}{dx}-3x+6x^2\frac{dy}{dx}
$

Lost! Where do I begin?!

get every term with dy/dx on the same side , then factor out the dy/dx

.

3. Thanks skeeter. A few questions,

For question #2, you said that there were 4 answers? I only see 2? I know that tangent has a periodicity of pi so my 2 answers are:

pi/3 and 4pi/3? Where are the other 2 solutions? If I add another period, I would get 7pi/3 which is greater than 2pie

Ok, I have tacked Question#3, but would like feedback on my algebra. This is me attempting to gather like terms ( I am a bit unsure if I am alllowed to minus the whole term $5xy\frac{dy}{dx}$ but I went ahead with it.... Should I have divided the left side by $5xy$ instead?

I hope what I said makes some sense.

$
-3xy^2\frac{dy}{dx}+\frac{dy}{dx}-5xy\frac{dy}{dx}-6x^2\frac{dy}{dx}=-1+6x^2-3x
$

Now, factor out the dy/dx..

$
\frac{dy}{dx}(-3xy^2+1-5xy-6x^2)=-1+6x^2-3x
$

4. you're looking for solutions in the interval $[0 , 2\pi]$

$\cot{x} = \frac{1}{\sqrt{3}}$ at $x = \frac{\pi}{3}$ and $x = \frac{4\pi}{3}$

$\cot{x} = -\frac{1}{\sqrt{3}}$ at $x = \frac{2\pi}{3}$ and $x = \frac{5\pi}{3}$

$
-3xy^2\frac{dy}{dx}+\frac{dy}{dx}-5xy\frac{dy}{dx}-6x^2\frac{dy}{dx}=-1+6x^2-3x
$

Now, factor out the dy/dx..

$
\frac{dy}{dx}(-3xy^2+1-5xy-6x^2)=-1+6x^2-3x
$
finish by dividing both sides by $(-3xy^2+1-5xy-6x^2)$

5. OMG YES! ARRGGHH!! I took the square root... hence the 4 solutions!