Help please! Methods 1/2

[Andiio]

5. Find the values of x between 0 and 2pi for which:

a) cosec x = 2
b) cot x = root3
c) secx + root2 = 0
d) cosec x = sec x

Could you please explain how to do them and show relevant working?

Much appreciated!

*I'd normally know how to do this but my teacher is absolutely useless ==

[taiga]

cosecx =2
hence 1/sinx = 2
Flip it all round then sinx = 1/2
Solve for that as per usual. (remember sinx can not equal 0)

for the second one, same logic
cotx = root3
hence tanx = 1/root3
solve as per usual

secx = -root2
1/cos = -root2
hence cos = -1/root2

last one is a bit more tricky

1/sinx = 1/cos x
hence, sinx = cos x
Therefore tanx = 1

the important thing to remember is that you can not forget the final equations came from, you can never have 1/0, hence you need to make sure your answer does not make it so that the initial condition results in 1/0. It almost always won't matter, but it is an important thing to remember, and will also aid you in drawing the graphs when you get to that later on.

Sorry I don't know how to use the equation writing thingo

[Russ]

Sorry I don't know how to use the equation writing thingo

Apparently it's [tex] brackets. I'm just as lost as you though

[Andiio]

cosecx =2
hence 1/sinx = 2
Flip it all round then sinx = 1/2
Solve for that as per usual. (remember sinx can not equal 0)

for the second one, same logic
cotx = root3
hence tanx = 1/root3
solve as per usual

secx = -root2
1/cos = -root2
hence cos = -1/root2

last one is a bit more tricky

1/sinx = 1/cos x
hence, sinx = cos x
Therefore tanx = 1

the important thing to remember is that you can not forget the final equations came from, you can never have 1/0, hence you need to make sure your answer does not make it so that the initial condition results in 1/0. It almost always won't matter, but it is an important thing to remember, and will also aid you in drawing the graphs when you get to that later on.

Sorry I don't know how to use the equation writing thingo

Thanks heaps! That's what i thought at first but i wasn't sure about the domain evident in the question
So to solve the, for example sinx = 1/2, you'd just make it x = sin^-1(1/2) right? which gives you an answer of x = 30

[taiga]

cosecx =2
hence 1/sinx = 2
Flip it all round then sinx = 1/2
Solve for that as per usual. (remember sinx can not equal 0)

for the second one, same logic
cotx = root3
hence tanx = 1/root3
solve as per usual

secx = -root2
1/cos = -root2
hence cos = -1/root2

last one is a bit more tricky

1/sinx = 1/cos x
hence, sinx = cos x
Therefore tanx = 1

the important thing to remember is that you can not forget the final equations came from, you can never have 1/0, hence you need to make sure your answer does not make it so that the initial condition results in 1/0. It almost always won't matter, but it is an important thing to remember, and will also aid you in drawing the graphs when you get to that later on.

Sorry I don't know how to use the equation writing thingo

Thanks heaps! That's what i thought at first but i wasn't sure about the domain evident in the question
So to solve the, for example sinx = 1/2, you'd just make it x = sin^-1(1/2) right? which gives you an answer of x = 30

Oh hang on, technically, yes you are correct, but that is if you are assuming degrees. Have you learnt of the Unit Circle at this stage?

Essentially 30 degree is pi/6. And that is in the first *quadrant*. Sin values are also positive in the 2nd quadrant, and to find equivalent values in the 2nd quadrant you would need to make it pi-original answer, which would make the 2nd answer 5pi/6

If you don't understand much of what I just said you should look up the unit circle in your text book, if you don't understand it try and clear it up with your teacher/tutor, or if worst comes worse I can help you out haha

[Andiio]

cosecx =2
hence 1/sinx = 2
Flip it all round then sinx = 1/2
Solve for that as per usual. (remember sinx can not equal 0)

for the second one, same logic
cotx = root3
hence tanx = 1/root3
solve as per usual

secx = -root2
1/cos = -root2
hence cos = -1/root2

last one is a bit more tricky

1/sinx = 1/cos x
hence, sinx = cos x
Therefore tanx = 1

the important thing to remember is that you can not forget the final equations came from, you can never have 1/0, hence you need to make sure your answer does not make it so that the initial condition results in 1/0. It almost always won't matter, but it is an important thing to remember, and will also aid you in drawing the graphs when you get to that later on.

Sorry I don't know how to use the equation writing thingo

Thanks heaps! That's what i thought at first but i wasn't sure about the domain evident in the question
So to solve the, for example sinx = 1/2, you'd just make it x = sin^-1(1/2) right? which gives you an answer of x = 30

Oh hang on, technically, yes you are correct, but that is if you are assuming degrees. Have you learnt of the Unit Circle at this stage?

Essentially 30 degree is pi/6. And that is in the first *quadrant*. Sin values are also positive in the 2nd quadrant, and to find equivalent values in the 2nd quadrant you would need to make it pi-original answer, which would make the 2nd answer 5pi/6

If you don't understand much of what I just said you should look up the unit circle in your text book, if you don't understand it try and clear it up with your teacher/tutor, or if worst comes worse I can help you out haha

Haha i kinda do understand; so if it was sinx = 1/2, what would you do to solve the question?

Teacher has been away a lot and really doesn't know how to teach so.. ==

[the.watchman]

Sorry I don't know how to use the equation writing thingo

Apparently it's [tex] brackets. I'm just as lost as you though

Haha, try http://vcenotes.com/forum/index.php/topic,3137.0.html and http://vcenotes.com/viki/index.php/Help:LaTeX

[taiga]

cosecx =2
hence 1/sinx = 2
Flip it all round then sinx = 1/2
Solve for that as per usual. (remember sinx can not equal 0)

for the second one, same logic
cotx = root3
hence tanx = 1/root3
solve as per usual

secx = -root2
1/cos = -root2
hence cos = -1/root2

last one is a bit more tricky

1/sinx = 1/cos x
hence, sinx = cos x
Therefore tanx = 1

the important thing to remember is that you can not forget the final equations came from, you can never have 1/0, hence you need to make sure your answer does not make it so that the initial condition results in 1/0. It almost always won't matter, but it is an important thing to remember, and will also aid you in drawing the graphs when you get to that later on.

Sorry I don't know how to use the equation writing thingo

Thanks heaps! That's what i thought at first but i wasn't sure about the domain evident in the question
So to solve the, for example sinx = 1/2, you'd just make it x = sin^-1(1/2) right? which gives you an answer of x = 30

Oh hang on, technically, yes you are correct, but that is if you are assuming degrees. Have you learnt of the Unit Circle at this stage?

Essentially 30 degree is pi/6. And that is in the first *quadrant*. Sin values are also positive in the 2nd quadrant, and to find equivalent values in the 2nd quadrant you would need to make it pi-original answer, which would make the 2nd answer 5pi/6

If you don't understand much of what I just said you should look up the unit circle in your text book, if you don't understand it try and clear it up with your teacher/tutor, or if worst comes worse I can help you out haha

Haha i kinda do understand; so if it was sinx = 1/2, what would you do to solve the question?

Teacher has been away a lot and really doesn't know how to teach so.. ==

Ah okay so when you have values of 1/2, 1/root2, root3, 1, 0, (and a few more which escape me), you can find exact values

best way is to draw a triangle for each one

one is half an equilateral triangle, and the other another with sides of 1.



So take  the triangle on the right for example. the 30 degree angle has a 1 opposite to it, and a 2 on the hypotenuse. Hence sin30 = 1/2.

but 30 degrees can also be put as pi/6 radians. Pi Radians is equal to 180 degrees, so just think of pi as 180 degrees, and dividing that by six gives you 30 degrees. And pi/6 is your first answer. You need to know about quadrants and stuff to find the other answers. Try Searching the "unit circle" in your book, in order to learn more about the quadrants, there is only so much I can explain on here

In all honesty you should probably know this stuff, if your teacher hasnt covered them it is a bit annoying =\ It really takes someone sitting there explaining it to you for you to understand properly. It's very important for you to understand this before commencing specialist maths and 3/4 methods

Have a look through your textbook

[iNerd]

I just learnt this today at tuition; I'm in Y10 studying Y11 Methods so yh..someone correct me if I screw this up!
So Sin(x) = 1/2 and we want to solve for values of x, correct?
NB: This isn't a proper question because there is no domain specified.
1. We need a reference angle; lets call it alpha.
alpha = Sin inverse of half
alpha = pi over 6.
Therefore the answers are pi over 6 + pi, pi over 6 + 2pi etc etc as there is no domain!

[taiga]

I just learnt this today at tuition; I'm in Y10 studying Y11 Methods so yh..someone correct me if I screw this up!
So Sin(x) = 1/2 and we want to solve for values of x, correct?
NB: This isn't a proper question because there is no domain specified.
1. We need a reference angle; lets call it alpha.
alpha = Sin inverse of half
alpha = pi over 6.
Therefore the answers are pi over 6 + pi, pi over 6 + 2pi etc etc as there is no domain!

trust mhs boys to be learning 1/2 in year 10

Erm yes you are almost right, but sin would be pi- pi/6 not pi+ pi/6

That said, I don't think your explanation will help the OP

[iNerd]

I just learnt this today at tuition; I'm in Y10 studying Y11 Methods so yh..someone correct me if I screw this up!
So Sin(x) = 1/2 and we want to solve for values of x, correct?
NB: This isn't a proper question because there is no domain specified.
1. We need a reference angle; lets call it alpha.
alpha = Sin inverse of half
alpha = pi over 6.
Therefore the answers are pi over 6 + pi, pi over 6 + 2pi etc etc as there is no domain!

trust mhs boys to be learning 1/2 in year 10

Erm yes you are almost right, but sin would be pi- pi/6 not pi+ pi/6

That said, I don't think your explanation will help the OP

ehh yh im a shit-head , thats what i meant