# solve cosec^2 2x - cot 2x = 1

• Aug 23rd 2012, 06:56 PM
Furyan
solve cosec^2 2x - cot 2x = 1
Hello

The question is to solve

$\csc^2 2x - \cot 2x = 1$

I have tried writing $\csc^2 2x$ in terms of $\cot^2 2x$

and $\cot 2x$ in terms of $\csc 2x$

without success.

When I caved in and looked at the mark scheme it says to using $\csc^2 2x \equiv 1 + \cot^2 2x$, which I didn't get. Could someone please give me a clue and or suggest another way.

Thank you
• Aug 23rd 2012, 07:19 PM
Prove It
Re: solve cosec^2 2x - cot 2x = 1
Quote:

Originally Posted by Furyan
Hello

The question is to solve

$\csc^2 2x - \cot 2x = 1$

I have tried writing $\csc^2 2x$ in terms of $\cot^2 2x$

and $\cot 2x$ in terms of $\csc 2x$

without success.

When I caved in and looked at the mark scheme it says to using $\csc^2 2x \equiv 1 + \cot^2 2x$, which I didn't get. Could someone please give me a clue and or suggest another way.

Thank you

\displaystyle \begin{align*} \csc^2{2x} - \cot{2x} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\cos{2x}}{\sin{2x}} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ \frac{1 - \sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ 1 - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \sin^2{2x} + \cos^2{2x} - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \cos^2{2x} - \sin{2x}\cos{2x} &= 0 \\ \cos{2x}\left( \cos{2x} - \sin{2x} \right) &= 0 \\ \cos{2x} = 0 \textrm{ or } \cos{2x} - \sin{2x} = 0 \end{align*}

Solve each of these equations :)
• Aug 23rd 2012, 07:42 PM
MaxJasper
Re: solve cosec^2 2x - cot 2x = 1
8 real roots between 0-2pi are:

$\left\{\left\{x\to \frac{\pi }{4}\right\},\left\{x\to \frac{3 \pi }{4}\right\},\left\{x\to \frac{5 \pi }{4}\right\},\left\{x\to \frac{7 \pi }{4}\right\},\left\{x\to \frac{9 \pi }{8}\right\},\left\{x\to \frac{\pi }{8}\right\},\left\{x\to \frac{5 \pi }{8}\right\},\left\{x\to \frac{13 \pi }{8}\right\}\right\}$

http://mathhelpforum.com/attachment....1&d=1345776091
• Aug 23rd 2012, 08:24 PM
SkyCapri
Re: solve cosec^2 2x - cot 2x = 1
Substitute 1+cot^2(2x) into the equation

Then we'll obtain cot^2(2x)-cot(2x)=0

Factorise the equation
(cot (2x))(cot (2x)-1)=0

cot 2x=0
1/ tan (2x)=0
For 1/tan (2x)=0,
tan (2x)= infinity
So 2x= 90, 270
x= 45, 135

cot (2x)-1=0
1/ tan (2x)=1
tan (2x)=1
2x=45, 225
x=22.5, 112.5
• Aug 23rd 2012, 08:56 PM
Prove It
Re: solve cosec^2 2x - cot 2x = 1
Quote:

Originally Posted by SkyCapri
Substitute 1+cot^2(2x) into the equation

Then we'll obtain cot^2(2x)-cot(2x)=0

Factorise the equation
(cot (2x))(cot (2x)-1)=0

cot 2x=0
1/ tan (2x)=0
For 1/tan (2x)=0,
tan (2x)= infinity
So 2x= 90, 270
x= 45, 135

cot (2x)-1=0
1/ tan (2x)=1
tan (2x)=1
2x=45, 225
x=22.5, 112.5

Don't forget that the cotangent function is periodic every \displaystyle \begin{align*} 180^{\circ} \end{align*}, so you also need to add all integer multiples of this.
• Aug 24th 2012, 05:21 AM
Furyan
Re: solve cosec^2 2x - cot 2x = 1
Quote:

Originally Posted by Prove It
\displaystyle \begin{align*} \csc^2{2x} - \cot{2x} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\cos{2x}}{\sin{2x}} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ \frac{1 - \sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ 1 - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \sin^2{2x} + \cos^2{2x} - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \cos^2{2x} - \sin{2x}\cos{2x} &= 0 \\ \cos{2x}\left( \cos{2x} - \sin{2x} \right) &= 0 \\ \cos{2x} = 0 \textrm{ or } \cos{2x} - \sin{2x} = 0 \end{align*}

Solve each of these equations :)

Thank you very much indeed. I had done exactly that.
Although I hadn't worked through to the solutions because I had it in my head that I had to get a quadratic in terms of a single trigonometric function.

It didn't occur to me that I could use $R\cos(\theta + \alpha)$.
It's good to know there's another approach.

Although I also now see why $\csc^2 2x \equiv \1 + cot^2 2x$

Dividing the identity $\sin^2 2x + \cos^2 2x = 1$ by $\sin^2 2x$.

I haven't actually worked through the problem yet, but hopefully I'll be able to now. I'll try both ways.

Thanks again :)