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Solve -2/3cot(2x)=0 Help

Solve -2/3cot(2x)=0 for x

For some random reason I keep getting 0 although there is two solutions in the range 0-pi

Help

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Reply 1
Avatar for Zacken
Zacken
22
Original post by Sniperdon227
Solve -2/3cot(2x)=0 for x

For some random reason I keep getting 0 although there is two solutions in the range 0-pi

Help


Do you mean -2/(3 cot (2x)) or (-2/3) (cot 2x)?
Original post by Zacken
Do you mean -2/(3 cot (2x)) or (-2/3) (cot 2x)?


(-2/3) (cot 2x)
Reply 3
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Zacken
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Original post by Sniperdon227
(-2/3) (cot 2x)


Your expression is zero precisely when cot⁡2x=cos⁡2xsin⁡2x=0⇒cos⁡2x=0\cot 2x = \frac{\cos 2x}{\sin 2x} = 0 \Rightarrow \cos 2x = 0.

Can you solve the final equation? cos⁡0=1\cos 0 = 1 so x=0x=0 isn't a solution.

Instead, what angle makes cos⁡θ=0\cos \theta = 0?
Original post by Zacken
Your expression is zero precisely when cot⁡2x=cos⁡2xsin⁡2x=0⇒cos⁡2x=0\cot 2x = \frac{\cos 2x}{\sin 2x} = 0 \Rightarrow \cos 2x = 0.

Can you solve the final equation? cos⁡0=1\cos 0 = 1 so x=0x=0 isn't a solution.

Instead, what angle makes cos⁡θ=0\cos \theta = 0?


Got it so cos(2x)=0 then solve for x rather weird but why cant i write it as
(-2/3) x (1/tan2x) =0 then solve with tan2x=0 i get all x values as 0 :frown:
Reply 5
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Zacken
22
Original post by Sniperdon227
Got it so cos(2x)=0 then solve for x rather weird but why cant i write it as
(-2/3) x (1/tan2x) =0 then solve with tan2x=0 i get all x values as 0 :frown:


But if tan 2x = 0 then you get (-2/3) / 0 and you can't divide by 0, can you...?

It's like saying 1/x = 0] so x=0x=0.
Original post by Sniperdon227
Solve -2/3cot(2x)=0 for x

For some random reason I keep getting 0 although there is two solutions in the range 0-pi

Help


i got pi/4 and 3pi/4
is that right?
Original post by Zacken
But if tan 2x = 0 then you get (-2/3) / 0 and you can't divide by 0, can you...?

It's like saying 1/x = 0] so x=0x=0.


-2cos(2x) / 3sin(2x) =0

-2cos(2x)=0

cos(2x)= 0/-2

same here aint it?
(edited 7 years ago)
Reply 8
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Zacken
22
Original post by Sniperdon227
-2cos(2x) / 3sin(2x) =0

-2cos(2x)=0

cos(2x)= 0/-2

same here aint it?


Yeah, cos 2x = 0/ -2 = 0.

There's a difference between a / 0 and 0 / a.

The former is undefined and the latter is 0.
Original post by Zacken
Yeah, cos 2x = 0/ -2 = 0.

There's a difference between a / 0 and 0 / a.

The former is undefined and the latter is 0.


Lool makes me question my math abilities thanks for your rapid responses
Reply 10
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Zacken
22
Original post by Sniperdon227
Lool makes me question my math abilities thanks for your rapid responses


No worry! :smile:
heres what i did,

because cot2x = 1/tan2x

(-2/3)(1/tan2x)=0

There is a double angle so, tan2x= 2tanx/1-tan^2x

(-2/3)(1/2tanx/1-tan^2x)=0
(-2/3)(1-tan^2x/2tanx)=0

multiply out

-2/3+2/3tan^2x (all divided by) 2tanx=0

multiply by 2tanx

-2/3+2/3tan^2x=0

factorise out

2/3(-1+tan^2x)=0

tan^2x=1

square root both sides

tanx=(plus or minus) 1

x=tan^-1(1) and x=tan^-1(-1) *remember cast diagrams, so do pi minus your answer to tan^-1(1)*

x= pi/4 and x=3pi/4
Original post by Sniperdon227
Solve -2/3cot(2x)=0 for x

For some random reason I keep getting 0 although there is two solutions in the range 0-pi

Help


where did you get this question from btw?
Original post by Alisahussain1
heres what i did,

because cot2x = 1/tan2x

(-2/3)(1/tan2x)=0

There is a double angle so, tan2x= 2tanx/1-tan^2x

(-2/3)(1/2tanx/1-tan^2x)=0
(-2/3)(1-tan^2x/2tanx)=0

multiply out

-2/3+2/3tan^2x (all divided by) 2tanx=0

multiply by 2tanx

-2/3+2/3tan^2x=0

factorise out

2/3(-1+tan^2x)=0

tan^2x=1

square root both sides

tanx=(plus or minus) 1

x=tan^-1(1) and x=tan^-1(-1) *remember cast diagrams, so do pi minus your answer to tan^-1(1)*

x= pi/4 and x=3pi/4



Thanks ive just done it your way to make sure I've got all angles covered for these type of questions, I got the correct answer using your method btw it was 3 marks
Reply 14
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Zacken
22
Original post by Alisahussain1
heres what i did,


That's a bit convoluted, really. Solving cos 2x = 0 is much easier.
Original post by Sniperdon227
Thanks ive just done it your way to make sure I've got all angles covered for these type of questions, I got the correct answer using your method btw it was 3 marks


no problem glad i could help!
it was probably 3 marks because that type of question may not have came up before and they dont want too many students to lose those marks, thats what i like to think anyway :biggrin:
was this from a past paper and if so which one?
Original post by Zacken
That's a bit convoluted, really. Solving cos 2x = 0 is much easier.


but you dont always get the correct answer and even if you did get the correct answer you might loose method marks
and i dont think using cos2x=0 would solve the equation correctly, would it?
Reply 17
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Zacken
22
Original post by Alisahussain1
but you dont always get the correct answer and even if you did get the correct answer you might loose method marks
and i dont think using cos2x=0 would solve the equation correctly, would it?


What? That would solve the equation correctly, of course. It was worth 3 marks because all you had to do was spot that you need to solve cos 2x = 0. Not what you did.

cot 2x = cos 2x / sin 2x.

So cot 2x = 0 means cos 2x = 0 * sin 2x = 0.
Original post by Alisahussain1
but you dont always get the correct answer and even if you did get the correct answer you might loose method marks
and i dont think using cos2x=0 would solve the equation correctly, would it?


Of course you always get the correct answer if the maths is correct....

And yes cos2x would solve it correctly and also save time
Original post by Zacken
What? That would solve the equation correctly, of course. It was worth 3 marks because all you had to do was spot that you need to solve cos 2x = 0. Not what you did.

cot 2x = cos 2x / sin 2x.

So cot 2x = 0 means cos 2x = 0 * sin 2x = 0.


im sorry but that makes no sense to me at all :s-smilie: but as long as you understand it thats fine with me

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