- #1

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## Homework Statement

Solve the following equation, where 0 [tex]\leq[/tex]u [tex]\leq[/tex] 2pi [360 degrees]

2sinucosu = 0

## Homework Equations

## The Attempt at a Solution

(2sinucosu/2) = (0/2)

sinucosu = 0

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- Thread starter Draggu
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- #1

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Solve the following equation, where 0 [tex]\leq[/tex]u [tex]\leq[/tex] 2pi [360 degrees]

2sinucosu = 0

(2sinucosu/2) = (0/2)

sinucosu = 0

- #2

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Try seeing if there's a specific trig identity to help you

- #3

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sin2u?

tried that as well, and not sure what to do.

tried that as well, and not sure what to do.

- #4

Mark44

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- #5

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## Homework Statement

Solve the following equation, where 0 [tex]\leq[/tex]u [tex]\leq[/tex] 2pi [360 degrees]

2sinucosu = 0

## Homework Equations

## The Attempt at a Solution

(2sinucosu/2) = (0/2)

sinucosu = 0

You gotta read this step by step alright no peeping into the next step.

Refer to the rule

Sin 2A=2 SinA CosA

0 [tex]\leq[/tex]u [tex]\leq[/tex] 2pi

0 [tex]\leq[/tex]2u [tex]\leq[/tex] 4pi

- #6

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going off of what he was saying, imagine for a second that the 2 wasn't there, so that you have sin(u)=0. For confusions sake, let's call it n now. So you've got sin(n)=0 Essentially saying the sign of some angle "n" equals 0 right? You should be able to figure out what that angle n is. If you think of it like that and can solve for n, then all you have to do now is think of n=2u, and that i think you can figure out =). Hope that wasn't too revealing =X

- #7

HallsofIvy

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So either sin u= 0 or cos u= 0. What values if y satisfy those?## Homework Statement

Solve the following equation, where 0 [tex]\leq[/tex]u [tex]\leq[/tex] 2pi [360 degrees]

2sinucosu = 0

## Homework Equations

## The Attempt at a Solution

(2sinucosu/2) = (0/2)

sinucosu = 0

- #8

- 446

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So either sin u= 0 or cos u= 0. What values if y satisfy those?

haha.. tats true we nid consider tat factor..

Therefore u = 2pi

- #9

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haha.. tats true we nid consider tat factor..

Therefore u = 2pi

Well.... I got as far as sinu = 0 . I honestly have no idea how to continue it further, since no special triangle can get you to equal 0. On a sin graph with a period of 2pi though, sin touches 0 at pi and 2pi

- #10

HallsofIvy

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Apparently youhaha.. tats true we nid consider tat factor..

Therefore u = 2pi

- #11

HallsofIvy

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Yes, exactly. And since your problem says "[itex]0\le u\le 2\pi[/itex], those are two solutions. Now, what values of u make cos(u)= 0?Well.... I got as far as sinu = 0 . I honestly have no idea how to continue it further, since no special triangle can get you to equal 0. On a sin graph with a period of 2pi though, sin touches 0 at pi and 2pi

- #12

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Yes, exactly. And since your problem says "[itex]0\le u\le 2\pi[/itex], those are two solutions. Now, what values of u make cos(u)= 0?

pi/2 and 3pi/2

I have a question though, how would I show my work for it? It's worth 6 marks...would I sketch the graph for each of them or? :)

- #13

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pi/2 and 3pi/2

I have a question though, how would I show my work for it? It's worth 6 marks...would I sketch the graph for each of them or? :)

Also, is using cosu necessary for the equation too? If so, changing it to sin2u = 0 is irrelevant. Since sin2u = 0 and diving sin2u by 2 and 0 by 2, giving us sinu = 0 , then would cos not be apart of this?

- #14

HallsofIvy

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No sin(au)/a is NOT equal to sin(u).

If xy= 0, then either x= 0 or y= 0. Surely you know that?

If xy= 0, then either x= 0 or y= 0. Surely you know that?

- #15

Mark44

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Also, is using cosu necessary for the equation too? If so, changing it to sin2u = 0 is irrelevant. Since sin2u = 0 and diving sin2u by 2 and 0 by 2, giving us sinu = 0 , then would cos not be apart of this?

You should get the same solutions either way:

By solving sin(2u) = 0

By solving sin(u)cos(u) = 0

- #16

HallsofIvy

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But NOT by "dividing sin 2u= 0 by 2" to get sin u= 0!

- #17

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No sin(au)/a is NOT equal to sin(u).

If xy= 0, then either x= 0 or y= 0. Surely you know that?

u=theta

sin2u=0 , is the double angle identity. Sorry but cos has nothing to do with the answer.

2pi/2 = pi pi is the period.

- #18

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No sin(au)/a is NOT equal to sin(u).

If xy= 0, then either x= 0 or y= 0. Surely you know that?

u=theta

sin2u=0 , is the double angle identity. Sorry but cos has nothing to do with the answer.

(2pi/2) = pi ---->pi is the period.

- #19

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u=theta

sin2u=0 , is the double angle identity. Sorry but cos has nothing to do with the answer.

(2pi/2) = pi ---->pi is the period.

Mark44 was right

[tex]sin(2u) = 0[/tex] is the same thing as [tex]sin(u)cos(u) = 0[/tex]. You will get the same answers either way, it is just a matter of preference.

- #20

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Hmm, well my answers were (pi/2), (3pi/2), 2pi, pi

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