- Thread starter
- #1

- Thread starter SarahJeen
- Start date

- Thread starter
- #1

- Admin
- #2

- Thread starter
- #3

could we divide both sides by tan?Let's do these one at a time for clarity. For the first problem, what do you think would be a good first step?

What if \(\displaystyle \displaystyle \tan{(x)} = 0\)?could we divide both sides by tan?

- Thread starter
- #5

if it was radians when sin(x) = 0 tan(x) = 0... Reasoning for that is that tan(x) = sin(x)/cos(x) so if sin(x) = 0 then tan(x) so sin(x) = at pi and 2pi... if i'm correct...What if \(\displaystyle \displaystyle \tan{(x)} = 0\)?

- - - Updated - - -

this is a very new topic for me and my final is next weekif it was radians when sin(x) = 0 tan(x) = 0... Reasoning for that is that tan(x) = sin(x)/cos(x) so if sin(x) = 0 then tan(x) so sin(x) = at pi and 2pi... if i'm correct...

- Admin
- #6

We could, but only if we consider \(\displaystyle \tan(x)=0\) is then possibly removed as a solution. A better approach would be to subtract $\tan(x)$ from both sides, then factor, so that we can use the zero factor property.could we divide both sides by tan?

So, subtracting, then factoring, what do we get?

- Thread starter
- #7

If someone could show the solutions for these 2 and also explain it I would appreicate it very muchif it was radians when sin(x) = 0 tan(x) = 0... Reasoning for that is that tan(x) = sin(x)/cos(x) so if sin(x) = 0 then tan(x) so sin(x) = at pi and 2pi... if i'm correct...

- - - Updated - - -

this is a very new topic for me and my final is next week

You seem to have missed my point entirely, which is that you can NOT divide by 0. Rather, you need to set your equation equal to 0, factorise, and set each factor equal to 0.if it was radians when sin(x) = 0 tan(x) = 0... Reasoning for that is that tan(x) = sin(x)/cos(x) so if sin(x) = 0 then tan(x) so sin(x) = at pi and 2pi... if i'm correct...

- - - Updated - - -

this is a very new topic for me and my final is next week

And no, the policy of MHB is to NOT give full solutions. Students are expected to be doing their own work, and so are required to attempt these problems themselves, showing their full working and where they are stuck. Then more specific guidance can be given.

The reason that Mark gave you a full solution in the other thread is because you posted on another site. Now that you are a member here you are expected to follow the rules of the forum.

- Thread starter
- #9

Ooo.. Thanks I will follow the guidelines I'am just struggling at this concept that is the reason for my....You seem to have missed my point entirely, which is that you can NOT divide by 0. Rather, you need to set your equation equal to 0, factorise, and set each factor equal to 0.

And no, the policy of MHB is to NOT give full solutions. Students are expected to be doing their own work, and so are required to attempt these problems themselves, showing their full working and where they are stuck. Then more specific guidance can be given.

The reason that Mark gave you a full solution in the other thread is because you posted on another site. Now that you are a member here you are expected to follow the rules of the forum.

- - - Updated - - -

Is this a good start? 2tanxcosx-tanx=0 ... tanx(2cosx-1)=0... so either tanx=0 or 2cosx-1=0Ooo.. Thanks I will follow the guidelines I'am just struggling at this concept that is the reason for my....

tanx=0 for x=k*Pi (k=1,2,...)

2cosx-1=0 ... 2cosx=1 ... cosx=1/2 ... x=Pi/3 + 2kPi and -Pi/3 +2kPi

- Admin
- #10

You will get much more from these if you are not simply shown how to do them, but actually take part in getting to the solutions.If someone could show the solutions for these 2 and also explain it I would appreicate it very much

We want to help you to be able to do these yourself. If you follow our suggestions you will get there in no time.

- Thread starter
- #11

Okay well if we subtract tan from both sidesYou will get much more from these if you are not simply shown how to do them, but actually take part in getting to the solutions.

We want to help you to be able to do these yourself. If you follow our suggestions you will get there in no time.

2tanxcosx-tanx

we would get

1tanxcosx? correct

Absolutely correct, though you should choose different letters for the different arbitrary constants. Well doneOoo.. Thanks I will follow the guidelines I'am just struggling at this concept that is the reason for my....

- - - Updated - - -

Is this a good start? 2tanxcosx-tanx=0 ... tanx(2cosx-1)=0... so either tanx=0 or 2cosx-1=0

tanx=0 for x=k*Pi (k=1,2,...)

2cosx-1=0 ... 2cosx=1 ... cosx=1/2 ... x=Pi/3 + 2kPi and -Pi/3 +2kPi

- Thread starter
- #13

I think I got it nowOkay well if we subtract tan from both sides

2tanxcosx-tanx

we would get

1tanxcosx? correct

2tanxcosx = tan

2tanxcosx - tanx = 0

tanx(2cosx - 1) = 0

tanx = 0, ===> x = 0, x=pi, and x=2pi >==========< ANSWER

Also,

2cosx -1 = 0, ===> cosx = 1/2

x = pi/3, and x=-pi/3 >====================< ANSWER

- Admin
- #14

Yes, that is a good start....

Is this a good start? 2tanxcosx-tanx=0 ... tanx(2cosx-1)=0... so either tanx=0 or 2cosx-1=0

tanx=0 for x=k*Pi (k=1,2,...)

2cosx-1=0 ... 2cosx=1 ... cosx=1/2 ... x=Pi/3 + 2kPi and -Pi/3 +2kPi

For \(\displaystyle \tan(x)=0\) you need to let $k$ be any integer, and then take only those values within the specified domain. Can you state which these are?

For \(\displaystyle \cos(x)=\frac{1}{2}\) which value(s) of $k$ do you want?

Recall we want \(\displaystyle 0\le x<2\pi\).

You are doing well, we just need to iron out these minor details.

- Thread starter
- #15

secound equation?

2sin^2x - sin x - 1 = 0?

Let m = sin x

2m^2 - m - 1 = 0

(m+1)(2m-1) = 0

m = -1

sin x = -1

x = 3pi/2

m = 1/2

sin x = 1/2

x = pi/6 or 5pi/6

- - - Updated - - -

2cosx = 1Yes, that is a good start.

For \(\displaystyle \tan(x)=0\) you need to let $k$ be any integer, and then take only those values withing the specified domain. Can you state which these are?

For \(\displaystyle \cos(x)=\frac{1}{2}\) which value(s) of $k$ do you want?

Recall we want \(\displaystyle 0\le x<2\pi\).

You are doing well, we just need to iron out these minor details.

cosx = .5

Is \(\displaystyle \displaystyle \begin{align*} -\frac{\pi}{3} \end{align*}\) in the region \(\displaystyle \displaystyle \begin{align*} x \in \left[ 0 , 2\pi \right] \end{align*}\)?I think I got it now

2tanxcosx = tan

2tanxcosx - tanx = 0

tanx(2cosx - 1) = 0

tanx = 0, ===> x = 0, x=pi, and x=2pi >==========< ANSWER

Also,

2cosx -1 = 0, ===> cosx = 1/2

x = pi/3, and x=-pi/3 >====================< ANSWER

- Thread starter
- #17

I put a new solution without the funny variblesYes, that is a good start.

For \(\displaystyle \tan(x)=0\) you need to let $k$ be any integer, and then take only those values within the specified domain. Can you state which these are?

For \(\displaystyle \cos(x)=\frac{1}{2}\) which value(s) of $k$ do you want?

Recall we want \(\displaystyle 0\le x<2\pi\).

You are doing well, we just need to iron out these minor details.

- - - Updated - - -

Yes ? am I correctIs \(\displaystyle \displaystyle \begin{align*} -\frac{\pi}{3} \end{align*}\) in the region \(\displaystyle \displaystyle \begin{align*} x \in \left[ 0 , 2\pi \right] \end{align*}\)?

- - - Updated - - -

Would you mind showing the correct solution now? since I have my input I would like to see how to do it properlyI put a new solution without the funny varibles

- - - Updated - - -

Yes ? am I correct

Last time I checked, there is not a single negative number that is greater than 0, so how could \(\displaystyle \displaystyle \begin{align*} x = -\frac{\pi}{3} \end{align*}\) possibly be in the region \(\displaystyle \displaystyle \begin{align*} x \in [ 0, 2\pi ] \end{align*}\)?

- Admin
- #19

While you have the right idea to recognize you have a quadratic in $\sin(x)$, you have factored incorrectly. Try FOILing your factored form...second equation?

2sin^2x - sin x - 1 = 0?

Let m = sin x

2m^2 - m - 1 = 0

(m+1)(2m-1) = 0

m = -1

sin x = -1

x = 3pi/2

m = 1/2

sin x = 1/2

x = pi/6 or 5pi/6