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# Division by zero multiple choice question watch

1. 4. Suppose that sin(2x) = 2cos(x). What can you conclude?
(a) x is of the form n + /2 where n is an integer
(b) x is of the form 2n + /2 where n is an integer
(c) x is of the form n where n is an integer
(d) x is of the form 2n where n is an integer
i've factorised sin2x=2cosx to 2cosx(sinx-1)=0
therefore x=arccos0 = 1, or x=arcsin1 = /2

from this I kind of guessed A by looking for an answer similar and it's right, i was just reading through the worked solutions and don't understand their conclusion fully, i believe it's something to do with general solutions, but i may possibly wrong.
2cosx(sinx-1)=0
implies: cosx=0, or sinx=1

x=(2n+1)/2 = + with n is a integer.
i dont understand the final line. 2n+1? cant think.
2. Remember sin(2x) = 2sin(x)cos(x).
3. (Original post by generalebriety)
Remember sin(2x) = 2sin(x)cos(x).
yes, i factorised it already, 2cosxsinx=2cosx,
2cosx(sinx-1)=0

please read post #1 again, last line in the 2nd quote i dont understand.

unless that relates to it and i just cant see it.
4. (Original post by KeineHeldenMehr)
yes, i factorised it already, 2cosxsinx=2cosx,
2cosx(sinx-1)=0

please read post #1 again, last line in the 2nd quote i dont understand.

unless that relates to it and i just cant see it.

instead of factorising, can you not instead divide by cosx, to get 2sinx=2 then sinx=1 and then its pretty easy from there
5. (Original post by stocker888)
instead of factorising, can you not instead divide by cosx, to get 2sinx=2 then sinx=1 and then its pretty easy from there
no no no, since cosx could = 0, potentially, you're losing a root.
6. You were on the right lines when you mentioned general solutions. Just think of where cosx = 0 (or sinx=1) from the graph. It doesn't just occur in one place.

If they haven't specified a particular range for the solutions they want, then you need to give the whole range. Here, the 2n+1 is for all integer values of n (so that translates to 1,3,5,7,... for n=0,1,2,3 resp. and similarly for all negative values of n).

Edit: Sorry, I should have explained why they got to the 2n+1 in more detail. notnek's post below does this.
7. (Original post by KeineHeldenMehr)
yes, i factorised it already, 2cosxsinx=2cosx,
2cosx(sinx-1)=0

please read post #1 again, last line in the 2nd quote i dont understand.

unless that relates to it and i just cant see it.
Think of the graphs of cos and sin. If cos(x)=0 then (for x>0), the first solution is x=(pi/2) and then every time you add pi the graph goes through the x axis so you have solutions of the form: x=(pi/2)+n*pi

If sin(x)=1 then the first solution is x=(pi/2) and then all other solutions will occur in periods of 2pi. So you have the solutions: x=(pi/2)+2n*pi

These solutions show that sin(x)=1 only if cos(x)=0 and all solutions are of the form x=(pi/2)+n*pi
8. (Original post by KeineHeldenMehr)
please read post #1 again, last line in the 2nd quote i dont understand.
Oh, right. All they've done from there is solved the equations cos x = 0 and sin x = 1, and lumped their answers together in one neat but uninformative form.
9. (Original post by m:)ckel)
You were on the right lines when you mentioned general solutions. Just think of where cosx = 0 (or sinx=1) from the graph. It doesn't just occur in one place.

If they haven't specified a particular range for the solutions they want, then you need to give the whole range. Here, the 2n+1 is for all integer values of n (so that translates to 1,3,5,7,... for n=0,1,2,3 resp. and similarly for all negative values of n).

Edit: Sorry, I should have explained why they got to the 2n+1 in more detail. notnek's post below does this.
(Original post by notnek)
Think of the graphs of cos and sin. If cos(x)=0 then (for x>0), the first solution is x=(pi/2) and then every time you add pi the graph goes through the x axis so you have solutions of the form: x=(pi/2)+n*pi

If sin(x)=1 then the first solution is x=(pi/2) and then all other solutions will occur in periods of 2pi. So you have the solutions: x=(pi/2)+2n*pi

These solutions show that sin(x)=1 only if cos(x)=0 and all solutions are of the form x=(pi/2)+n*pi
thanks guys. i was just thinking they both have a period of 2pi and was just wondering why it isnt x=(pi/2)+2n*pi since it does cross the x-axis during those points, doesnt it?
10. (Original post by KeineHeldenMehr)
thanks guys. i was just thinking they both have a period of 2pi and was just wondering why it isnt x=(pi/2)+2n*pi since it does cross the x-axis during those points, doesnt it?
cos(x) crosses the x-axis at every integer multiple of pi (see first line of notnek's post)

Remember that you find the period by observing when the curve starts to repeat itself. So the period for both is 2*pi, yes, but you need to find out when cos is 0, which is every pi units. Look at the graph if you are not sure..
11. (Original post by m:)ckel)
cos(x) crosses the x-axis at every integer multiple of pi (see first line of notnek's post)

Remember that you find the period by observing when the curve starts to repeat itself. So the period for both is 2*pi, yes, but you need to find out when cos is 0, which is every pi units. Look at the graph if you are not sure..
yeah, was thinking that too, wasnt too sure how it satisfies sinx=1 at the same time.
13. this is because all values of sinx=1 satisfy that of cosx=0. cosx=0 gives 0.5pi, 1.5pi, 2.5pi, 3.5pi....
14. rrrrrrrrrep me
sin(x) cos(x)=cos(x)
you are able to cancel the cos(x)'s without losing a root. suppose cos(x) were 0 then it reduces to 0=0 which is trivial, right?
so you're solving sin(x)=1.
pi/2,3pi/2,5pi/2,...etc
all have the form of (a) npi+pi/2
16. (Original post by latentcorpse)
sin(x) cos(x)=cos(x)
you are able to cancel the cos(x)'s without losing a root. suppose cos(x) were 0 then it reduces to 0=0 which is trivial, right?
so you're solving sin(x)=1.
pi/2,3pi/2,5pi/2,...etc
all have the form of (a) npi+pi/2
true, hmm.. that flew completely over my head.
thanks.

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