# trig equation - all solutions

• Jan 21st 2010, 12:03 PM
Tweety
trig equation - all solutions
solve $cosec^{2}2x - cot2x = 1$

$0 \leq x \leq 180$

replace $cosex^{2}2x$ with $1 + cot^{2}2x = cosex^{2}2x$

$1 + cot^{2}2x -cot2x = 1$

$cot^{2}2x -cot2x = 0$

$cot2x(cot2x-1) = 0$

$cot2x = 0$ or

$cot2x = 1$

for $cot2x =0$ is it correct to say there are no solutions for this?

for the other one I got $tan2x = 1$

$2x = 45, 225$

$x = 22.5 , 112.5$

Are these all the solutions in the interval, or does cot2x =0 also have a solution?
• Jan 21st 2010, 12:26 PM
bigwave
Quote:

Originally Posted by Tweety
solve $cosec^{2}2x - cot2x = 1$

$0 \leq x \leq 180$

replace $cosex^{2}2x$ with $1 + cot^{2}2x = cosex^{2}2x$

$1 + cot^{2}2x -cot2x = 1$

$cot^{2}2x -cot2x = 0$

$cot2x(cot2x-1) = 0$

$cot2x = 0$ or

$cot2x = 1$

for $cot2x =0$ is it correct to say there are no solutions for this?

for the other one I got $tan2x = 1$

$2x = 45, 225$

$x = 22.5 , 112.5$

Are these all the solutions in the interval, or does cot2x =0 also have a solution?

on $\cot{2x} = 0$

$\cot^{-1}{(0)} = \frac{\pi}{2}$

therefor

$2x = \frac{\pi}{2}$

so $x = \frac{\pi}{4}$

for $\pi \ge x \ge 0$
• Jan 21st 2010, 12:27 PM
The numerator of cot2x is cos2x.
This is zero for

$2x=\frac{\pi}{2}\ and\ \frac{3\pi}{2}$
• Jan 21st 2010, 12:48 PM
Tweety
Quote:

The numerator of cot2x is cos2x.
This is zero for

$2x=\frac{\pi}{2}\ and\ \frac{3\pi}{2}$

Thanks,

I have $cot2x = \frac{sin2x}{cos2x} = 0$

wouldn't I have to solve for x? so I would have to multiply both sides my cos2x to get $sin2x = 0$ , which is zero at $\pi$ and $2\pi$?

And so there are four solutions altogether?
• Jan 21st 2010, 12:57 PM
bigwave
cotx = cosx\sinx
$\cot\theta = \frac{cos\theta}{sin\theta}$
• Jan 21st 2010, 01:00 PM
Tweety
Quote:

Originally Posted by bigwave
as mentioned by Archie

$cot2x = \frac{1}{cos2x} = 0$

oh right, Can you please explain how did you get cot2x to equal $\frac{1}{cos2x}$ ?

Thank you!
• Jan 21st 2010, 01:02 PM
That one is the ratio for Tan2x, Tweety, ie $\frac{Sin2x}{Cos2x}$

$Cot2x=\frac{Cos2x}{Sin2x}$

$Cot2x=\frac{1}{Tan2x}=\frac{1}{\frac{Sin2x}{Cos2x} }$

which is Tan2x inverted.

So the other two solutions are found by solving Cos2x=0,
since Cot2x=0 when the numerator Cos2x is zero.
• Jan 21st 2010, 01:03 PM
Tweety
Actually I was just wondering,

I have $cot2x = 0$

that would be the same as witting $\frac{1}{tan2x} = 0$

...meaning tan2x couldn't = 0 ? So how come these other to solutions are 'valid'?

• Jan 21st 2010, 01:05 PM
You may be thinking of Tan2x going to infinity,
but yes it does when Cos2x goes to zero.

$\frac{1}{x}$ goes to zero when x goes to infinity.
• Jan 21st 2010, 01:06 PM
bigwave
Quote:

Originally Posted by Tweety
oh right, Can you please explain how did you get cot2x to equal $\frac{1}{cos2x}$ ?

Thank you!

sorry this was not correct...

• Jan 21st 2010, 01:10 PM
Tweety
Okay, I kind of 'get it' , thanks for your help.
• Jan 21st 2010, 01:23 PM
$Cot2x=\frac{1}{Tan2x}$

$Tan2x=\frac{Sin2x}{Cos2x}$

$\frac{1}{\frac{Sin2x}{Cos2x}}=\frac{\frac{Sin2x}{C os2x}\frac{Cos2x}{Sin2x}}{\frac{Sin2x}{Cos2x}}=\fr ac{\frac{Sin2x}{Cos2x}}{\frac{Sin2x}{Cos2x}}\frac{ Cos2x}{Sin2x}=\frac{Cos2x}{Sin2x}$

You can think it through with little fractions first.

$\frac{1}{2}=0.5$

$\frac{1}{1}=1$

$\frac{1}{0.5}=\frac{1}{\frac{1}{2}}=\frac{(2)\frac {1}{2}}{\frac{1}{2}}=2$

since the numerator is twice the denominator.

To divide by a fraction, just turn it upside down and multiply instead.

$\frac{\frac{3}{4}}{\frac{2}{5}}=\frac{3}{4}\frac{5 }{2}=\frac{15}{8}$

Therefore

$Cot2x=\frac{1}{Tan2x}$

is Tan2x upside down.