Solve Algebraic Equations 0≤x<2π | Exact Values

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Discussion Overview

The discussion revolves around solving algebraic equations involving trigonometric functions, specifically focusing on the equations 2tan(x)cos(x) = tan(x) and 2sin²(x) - sin(x) - 1 = 0. Participants are asked to provide exact values for solutions within the interval 0 ≤ x < 2π.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant suggests dividing both sides of the first equation by tan(x), raising the concern about what happens if tan(x) = 0.
  • Another participant discusses the implications of tan(x) = sin(x)/cos(x) and identifies points where sin(x) = 0, leading to potential solutions at x = π and x = 2π.
  • There is a suggestion to subtract tan(x) from both sides and factor the equation instead of dividing by tan(x) to avoid losing solutions.
  • Participants explore the factorization of the first equation, leading to the solutions tan(x) = 0 and 2cos(x) - 1 = 0, with discussions on the values of x that satisfy these equations.
  • For the second equation, one participant proposes letting m = sin(x) and factors it to find solutions for sin(x) = -1 and sin(x) = 1/2, identifying corresponding x values.
  • There are multiple confirmations of the correctness of the steps taken, with encouragement to clarify the values of k for the solutions within the specified domain.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the equations but express varying opinions on specific steps, such as whether to divide by tan(x) or to factor the equations. The discussion remains unresolved on the best method to proceed without losing potential solutions.

Contextual Notes

Participants note the importance of not dividing by zero and emphasize the need to set equations to zero and factor them. There are also discussions about the correct interpretation of arbitrary constants in the solutions.

Who May Find This Useful

This discussion may be useful for students preparing for exams in trigonometry or algebra, particularly those seeking to understand the process of solving trigonometric equations and the implications of different algebraic manipulations.

SarahJeen
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Solve the following equations algebraically. Give solutions as exact values where 0≤x<2π

Solve the following equations algeraically. Give solutions as exact values where 0≤x<2π

a)2tanxcosx=tanx
b) View attachment 862

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Any Help would be much appreciated thanks sarah :) (Inlove)
 

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Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x<2π

Let's do these one at a time for clarity. For the first problem, what do you think would be a good first step?
 
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x<2π

MarkFL said:
Let's do these one at a time for clarity. For the first problem, what do you think would be a good first step?

could we divide both sides by tan?
 
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x<2π

SarahJeen said:
could we divide both sides by tan?

What if [math]\displaystyle \tan{(x)} = 0[/math]?
 
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

Prove It said:
What if [math]\displaystyle \tan{(x)} = 0[/math]?

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...

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SarahJeen said:
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...

this is a very new topic for me and my final is next week (Sun)
 
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x<2π

SarahJeen said:
could we divide both sides by tan?

We could, but only if we consider $$\tan(x)=0$$ 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.

So, subtracting, then factoring, what do we get?
 
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

SarahJeen said:
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 (Sun)

If someone could show the solutions for these 2 and also explain it I would appreicate it very much :) (Bandit)
 
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

SarahJeen said:
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 (Sun)

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.
 
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&amp;lt;2π

Prove It said:
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.

Ooo.. Thanks :) I will follow the guidelines I'am just struggling at this concept that is the reason for my...

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SarahJeen said:
Ooo.. Thanks :) I will follow the guidelines I'am just struggling at this concept that is the reason for my...

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
 
  • #10
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

SarahJeen said:
If someone could show the solutions for these 2 and also explain it I would appreicate it very much :) (Bandit)

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.

We want to help you to be able to do these yourself. If you follow our suggestions you will get there in no time. :D
 
  • #11
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

MarkFL said:
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.

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

Okay :) well if we subtract tan from both sides

2tanxcosx-tanx

we would get

1tanxcosx? correct
 
  • #12
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&amp;lt;2π

SarahJeen said:
Ooo.. 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

Absolutely correct, though you should choose different letters for the different arbitrary constants. Well done :)
 
  • #13
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

SarahJeen said:
Okay :) well if we subtract tan from both sides

2tanxcosx-tanx

we would get

1tanxcosx? correct

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
 
  • #14
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&amp;lt;2π

SarahJeen said:
...
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

Yes, that is a good start.

For $$\tan(x)=0$$ 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 $$\cos(x)=\frac{1}{2}$$ which value(s) of $k$ do you want?

Recall we want $$0\le x<2\pi$$.

You are doing well, we just need to iron out these minor details. :D
 
  • #15
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

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

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MarkFL said:
Yes, that is a good start.

For $$\tan(x)=0$$ 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 $$\cos(x)=\frac{1}{2}$$ which value(s) of $k$ do you want?

Recall we want $$0\le x<2\pi$$.

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

2cosx = 1
cosx = .5
 
  • #16
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

SarahJeen said:
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

Is [math]\displaystyle \begin{align*} -\frac{\pi}{3} \end{align*}[/math] in the region [math]\displaystyle \begin{align*} x \in \left[ 0 , 2\pi \right] \end{align*}[/math]?
 
  • #17
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&amp;amp;amp;lt;2π

MarkFL said:
Yes, that is a good start.

For $$\tan(x)=0$$ 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 $$\cos(x)=\frac{1}{2}$$ which value(s) of $k$ do you want?

Recall we want $$0\le x<2\pi$$.

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

I put a new solution without the funny varibles

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Prove It said:
Is [math]\displaystyle \begin{align*} -\frac{\pi}{3} \end{align*}[/math] in the region [math]\displaystyle \begin{align*} x \in \left[ 0 , 2\pi \right] \end{align*}[/math]?

Yes ? am I correct

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SarahJeen said:
I put a new solution without the funny varibles

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Yes ? am I correct

Would you mind showing the correct solution now? since I have my input I would like to see how to do it properly
 
  • #18
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x<2π

Last time I checked, there is not a single negative number that is greater than 0, so how could [math]\displaystyle \begin{align*} x = -\frac{\pi}{3} \end{align*}[/math] possibly be in the region [math]\displaystyle \begin{align*} x \in [ 0, 2\pi ] \end{align*}[/math]?
 
  • #19
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

SarahJeen said:
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

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

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