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

Hello

The question is to solve

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

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

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

without success.

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

Thank you

Re: solve cosec^2 2x - cot 2x = 1

Quote:

Originally Posted by

**Furyan** Hello

The question is to solve

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

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

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

without success.

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

Thank you

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

Solve each of these equations :)

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Re: solve cosec^2 2x - cot 2x = 1

8 real roots between 0-2pi are:

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

http://mathhelpforum.com/attachment....1&d=1345776091

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

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 $\displaystyle \displaystyle \begin{align*} 180^{\circ} \end{align*}$, so you also need to add all integer multiples of this.

Re: solve cosec^2 2x - cot 2x = 1

Quote:

Originally Posted by

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

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 $\displaystyle R\cos(\theta + \alpha)$.

It's good to know there's another approach.

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

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

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

Thanks again :)