Someone wanna check my math?

[TheRyuu]

I'm not asking how to do a problem, just if I did it right.
Ok, it may be a long one:

(for the purpose of writing on the computer, # is gonna be pie (3.14), and it would be like your typing it into a calculator since there's only so much you can type)
Original question: "Find all solutions in the interval [0,2#): 2sin^2(x/4)-3cos(x/4)=0"

Ok, this is what I got:


THIS IS WHAT I MEANT by ALL of the sin^2(u) (in the pictures case x)

2sin^2(x/4)-3cos(x/4)=0

Let u=x/4

2sin^2(u)-3cos(u)

2(1-cos^2(u))-3cos(u)

2-2cos^2(u)-3cosu

(-1)-2cos^2(u)-3cos(u)+2

2cos^2(u)+3cos(u)-2
Factored:
(2cos(u)-1)(cos(u)+2)

Solving for each one:
2cos(u)-1=0
cos(u)=1/2
arccos(1/2)=u
u=#/3, 5#/3

And the other:
cosu+2=0
cosu=-2
arccos(-2)=u
Doesn't work so I reject this answer?

Anyway, so (substitute back in):
x/4=#/3 , x/4=5#/3

so:
x=4#/3 , x=20#/3

I'm only solving for [0,2#) so:
{x=4#/3}

And that would be my final answer?
Just wondering. I think I might have done something wrong since only 1 out of the 4 possible answers would be the correct one.

Edit:
Fixed stupid u^2

Edit2:
Fixed ^2 AGAIN.

 

[tasmanian]

Originally posted by: wizboy11
I'm not asking how to do a problem, just if I did it right.
Ok, it may be a long one:

(for the purpose of writing on the computer, # is gonna be pie (3.14), and it would be like your typing it into a calculator since there's only so much you can type)
Original question: "Find all solutions in the interval [0,2#): 2sin(x/4^2)-3cos(x/4)=0"

Ok, this is what I got:

2sin^2(x/4)-3cos(x/4)=0

Let u=x/4

2sin(u^2)-3cos(u)

2(1-cos(u^2))-3cos(u)

2-2cos(u^2)-3cosu

(-1)-2cos(u^2)-3cos(u)+2

2cos(u^2)+3cos(u)-2
Factored:
(2cos(u)-1)(cos(u)+2)

Solving for each one:
2cos(u)-1=0
cos(u)=1/2
arccos(1/2)=u
u=#/3, 5#/3

And the other:
cosu+2=0
cosu=-2
arccos(-2)=u
Doesn't work so I reject this answer?

Anyway, so:
x/4=#/3 , x/4=5#/3

so:
x=4#/3 , x=20#/3

I'm only solving for [0,2#) so:
{x=4#/3}

And that would be my final answer?
Just wondering. I think I might have done something wrong since only 1 out of the 4 possible answers would be the correct one.

You lost me after "this is what i got"

 

[TheRyuu]

Originally posted by: tasmanian
You lost me after "this is what i got"

lol

Come on, this is Pre-Calc stuff, easy stuff.

I'm probably not gonna get an answer since this is ATOT, just wanted to see if someone would reply.

I don't like those "help me with my homework" threads but they normally just have a question.
I actually did out the whole fsck'n problem.

Just wondering if I did something wrong.

 

[Spongo]

you lost me after the "original question."

 

[chuckywang]

2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0

(2cos(x/4)+1)(cos(x/4)-2) = 0

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2

So therefore x/4 = 2pi/3, 4pi/3, ....

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.

 

[TheRyuu]

Originally posted by: chuckywang
2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2
Did That

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0
I believe thats what I got

(2cos(x/4)+1)(cos(x/4)-2) = 0
How come I got (2cos(x/4)-1)(cos(x/4)+2), isn't that were the + and - should be? Unless I screwed it up.

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2
Were did the negative come from, I got +1/2

So therefore x/4 = 2pi/3, 4pi/3, ....
True for -1/2

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.
See above

 

[TuxDave]

Originally posted by: chuckywang
2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0

(2cos(x/4)+1)(cos(x/4)-2) = 0

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2

So therefore x/4 = 2pi/3, 4pi/3, ....

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.

:thumbsup: Much easier to read

Edit: Good point, factoring looks wrong

 

[tikwanleap]

I get 4*pi/3

btw, i'm sure it's just a typo but you keep switching between

2sin(x/4^2)

and

2sin^2(x/4)

which was really confusing at first...

But from following your steps, it looks like really meant sin^2(u) instead of sin(u^2) which are two really different functions.

 

[TheRyuu]

Originally posted by: tikwanleap
I get 4*pi/3

btw, i'm sure it's just a typo but you keep switching between

2sin(x/4^2)

and

2sin^2(x/4)

which was really confusing at first...

But from following your steps, it looks like really meant sin^2(u) instead of sin(u^2) which are two really different functions.

I can't do the superscript 2 after the sin
So, I probably meant sin(u^2)

And I think I fixed them all to (u^2)

 

[TuxDave]

Originally posted by: tikwanleap
I get 4*pi/3

btw, i'm sure it's just a typo but you keep switching between

2sin(x/4^2)

and

2sin^2(x/4)

which was really confusing at first...

But from following your steps, it looks like really meant sin^2(u) instead of sin(u^2) which are two really different functions.

My math teachers would deduct points if we ever wrote sin(x^2) instead of sin^2(x)

 

[TuxDave]

Originally posted by: wizboy11

Originally posted by: tikwanleap
I get 4*pi/3

btw, i'm sure it's just a typo but you keep switching between

2sin(x/4^2)

and

2sin^2(x/4)

which was really confusing at first...

But from following your steps, it looks like really meant sin^2(u) instead of sin(u^2) which are two really different functions.

I can't do the superscript 2 after the sin
So, I probably meant sin(u^2)

So wait a minute....

sin(u^2) in the problem above really meant sin(u*u) and not sin(u)*sin(u)?

Then how in the world did you use the identity.

sin(u^2) = 1-cos(u^2)

 

[TheRyuu]

Originally posted by: TuxDave

Originally posted by: tikwanleap
I get 4*pi/3

btw, i'm sure it's just a typo but you keep switching between

2sin(x/4^2)

and

2sin^2(x/4)

which was really confusing at first...

But from following your steps, it looks like really meant sin^2(u) instead of sin(u^2) which are two really different functions.

My math teachers would deduct points if we ever wrote sin(x^2) instead of sin^2(x)

Originally posted by: TuxDave
So wait a minute....

sin(u^2) in the problem above really meant sin(u*u) and not sin(u)*sin(u)?

Then how in the world did you use the identity.

sin(u^2) = 1-cos(u^2)

THIS IS WHAT I MEANT by ALL of the u^2 (in the pictures case x)

 

[tikwanleap]

Originally posted by: wizboy11

Originally posted by: chuckywang
2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2
Did That

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0
I believe thats what I got

(2cos(x/4)+1)(cos(x/4)-2) = 0
How come I got (2cos(x/4)-1)(cos(x/4)+2), isn't that were the + and - should be? Unless I screwed it up.

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2
Were did the negative come from, I got +1/2

So therefore x/4 = 2pi/3, 4pi/3, ....
True for -1/2

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.
See above

There is a mistake in chuckywang's step 3:

(2cos(x/4)+1)(cos(x/4)-2) = 0

if you multiply it back together you get:

2cos^2(x/4) - 3cos(x/4) - 2 = 0

multiply by -1 and rearrange the terms

2 - 2cos^2(x/4) + 3cos(x/4) = 0

which is not the same as chuckywang's step 2:

2 - 2cos(x/4)^2 - 3cos(x/4) = 0

 

[TuxDave]

Originally posted by: wizboy11

Originally posted by: TuxDave

Originally posted by: tikwanleap
I get 4*pi/3

btw, i'm sure it's just a typo but you keep switching between

2sin(x/4^2)

and

2sin^2(x/4)

which was really confusing at first...

But from following your steps, it looks like really meant sin^2(u) instead of sin(u^2) which are two really different functions.

My math teachers would deduct points if we ever wrote sin(x^2) instead of sin^2(x)

Originally posted by: TuxDave
So wait a minute....

sin(u^2) in the problem above really meant sin(u*u) and not sin(u)*sin(u)?

Then how in the world did you use the identity.

sin(u^2) = 1-cos(u^2)

THIS IS WHAT I MEANT by ALL of the u^2 (in the pictures case x)

-1 point for you.

 

[tikwanleap]

Originally posted by: wizboy11

Originally posted by: TuxDave

Originally posted by: tikwanleap
I get 4*pi/3

btw, i'm sure it's just a typo but you keep switching between

2sin(x/4^2)

and

2sin^2(x/4)

which was really confusing at first...

But from following your steps, it looks like really meant sin^2(u) instead of sin(u^2) which are two really different functions.

My math teachers would deduct points if we ever wrote sin(x^2) instead of sin^2(x)

Originally posted by: TuxDave
So wait a minute....

sin(u^2) in the problem above really meant sin(u*u) and not sin(u)*sin(u)?

Then how in the world did you use the identity.

sin(u^2) = 1-cos(u^2)

THIS IS WHAT I MEANT by ALL of the u^2 (in the pictures case x)

So you should type it like this:

sin^2(x)

 

[TheRyuu]

Originally posted by: TuxDave
-1 point for you.

Originally posted by: tikwanleap


So you should type it like this:

sin^2(x)

Look, If you write sin(x^2) in the calculator, it would be the same as in the picture.
Thats why I said it's as if I was writing it in a calculator. .

And I guess I'm right since no one has proved me wrong yet.

 

[TuxDave]

Originally posted by: wizboy11

Originally posted by: TuxDave
-1 point for you.

Look, If you write sin(x^2) in the calculator, it would be the same as in the picture.
Thats why I said it's as if I was writing it in a calculator.

Do you own a funky HP Calculator that you have type things backwards to get them to add? My TI-89 if I typed sin(x^2) it gives me sin(x*x)

 

[tikwanleap]

Originally posted by: wizboy11

Originally posted by: TuxDave
-1 point for you.

Originally posted by: tikwanleap


So you should type it like this:

sin^2(x)

Look, If you write sin(x^2) in the calculator, it would be the same as in the picture.
Thats why I said it's as if I was writing it in a calculator.

true, but then you should write it like this.

sin(x)^2

not

sin(x^2)

The second one means to square the x first and then apply the sin function, which is totally wrong in this problem.

 

[TheRyuu]

Originally posted by: tikwanleap

Originally posted by: wizboy11

Originally posted by: TuxDave
-1 point for you.

Originally posted by: tikwanleap


So you should type it like this:

sin^2(x)

Look, If you write sin(x^2) in the calculator, it would be the same as in the picture.
Thats why I said it's as if I was writing it in a calculator.

true, but then you should write it like this.

sin(x)^2

not

sin(x^2)

The second one means to square the x first and then doing the sin function, which is totally wrong in this problem.

Then read the problem the right way then and tell me I'm right.
I'm not re-writing the whole dam thing so "pretend" that it's sin(x)^2.

 

[chuckywang]

Originally posted by: wizboy11

Originally posted by: chuckywang
2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2
Did That

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0
I believe thats what I got

(2cos(x/4)+1)(cos(x/4)-2) = 0
How come I got (2cos(x/4)-1)(cos(x/4)+2), isn't that were the + and - should be? Unless I screwed it up.

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2
Were did the negative come from, I got +1/2

So therefore x/4 = 2pi/3, 4pi/3, ....
True for -1/2

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.
See above

Crap, you're right. I was doing my math too fast and I didn't get all the signs down correctly.

It should be cos(x/4) = 1/2.

Therefore x=4pi/3

 

[tikwanleap]

Anyways the standard way of typing it is:

sin^2(x)

Then people will know to square after doing the sin function instead of before inside the paranthesis.

Don't forget that parenthesis denotes order of operation with the inner most parenthesis evaluated first.

 

[TheRyuu]

Originally posted by: chuckywang

Originally posted by: wizboy11

Originally posted by: chuckywang
2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2
Did That

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0
I believe thats what I got

(2cos(x/4)+1)(cos(x/4)-2) = 0
How come I got (2cos(x/4)-1)(cos(x/4)+2), isn't that were the + and - should be? Unless I screwed it up.

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2
Were did the negative come from, I got +1/2

So therefore x/4 = 2pi/3, 4pi/3, ....
True for -1/2

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.
See above

Crap, you're right. I was doing my math too fast and I didn't get all the signs down correctly.

It should be cos(x/4) = 1/2.

Therefore x=4pi/3

Yay, guess I'm right then
Just wanted to check. Thanks.

 

[tikwanleap]

Originally posted by: wizboy11

Originally posted by: chuckywang

Originally posted by: wizboy11

Originally posted by: chuckywang
2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2
Did That

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0
I believe thats what I got

(2cos(x/4)+1)(cos(x/4)-2) = 0
How come I got (2cos(x/4)-1)(cos(x/4)+2), isn't that were the + and - should be? Unless I screwed it up.

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2
Were did the negative come from, I got +1/2

So therefore x/4 = 2pi/3, 4pi/3, ....
True for -1/2

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.
See above

Crap, you're right. I was doing my math too fast and I didn't get all the signs down correctly.

It should be cos(x/4) = 1/2.

Therefore x=4pi/3

Yay, guess I'm right then
Just wanted to check. Thanks.

I posted that I got the same answer earlier.

 

[TuxDave]

Originally posted by: tikwanleap
I posted that I got the same answer earlier.

But the thread would've been less entertaining if it stopped there.

 

[TheRyuu]

Originally posted by: tikwanleap

Originally posted by: wizboy11

Originally posted by: chuckywang

Originally posted by: wizboy11

Originally posted by: chuckywang
2sin(x/4)^2-3cos(x/4)=0

Use the fact that sin(x)^2 = 1-cos(x)^2
Did That

Therefore, 2-2cos(x/4)^2 - 3cos(x/4) = 0
I believe thats what I got

(2cos(x/4)+1)(cos(x/4)-2) = 0
How come I got (2cos(x/4)-1)(cos(x/4)+2), isn't that were the + and - should be? Unless I screwed it up.

We all know that cos(x/4) can never equal 2.

Therefore, the only other option is cos(x/4) = -1/2
Were did the negative come from, I got +1/2

So therefore x/4 = 2pi/3, 4pi/3, ....
True for -1/2

Since x has to be between 0 and 2pi, the only solution is x=8pi/3.
See above

Crap, you're right. I was doing my math too fast and I didn't get all the signs down correctly.

It should be cos(x/4) = 1/2.

Therefore x=4pi/3

Yay, guess I'm right then
Just wanted to check. Thanks.

I posted that I got the same answer earlier.

ok then, great.

Right after I finally (hopefully) fixed the OP.

I guess it's time to let another thread die....

Good Night :moon: