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

#### SarahJeen

##### New member
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)

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Any Help would be much appreciated thanks sarah

#### MarkFL

Staff member
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?

#### SarahJeen

##### New member
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?
could we divide both sides by tan?

#### Prove It

##### Well-known member
MHB Math Helper
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x<2π

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

#### SarahJeen

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

What if $$\displaystyle \displaystyle \tan{(x)} = 0$$?
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...

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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

#### MarkFL

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

could we divide both sides by tan?
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.

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

#### SarahJeen

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

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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this is a very new topic for me and my final is next week
If someone could show the solutions for these 2 and also explain it I would appreicate it very much

#### Prove It

##### Well-known member
MHB Math Helper
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

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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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.

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.

#### SarahJeen

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

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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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

#### MarkFL

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

If someone could show the solutions for these 2 and also explain it I would appreicate it very much
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.

#### SarahJeen

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

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.
Okay well if we subtract tan from both sides

2tanxcosx-tanx

we would get

1tanxcosx? correct

#### Prove It

##### Well-known member
MHB Math Helper
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&amp;lt;2π

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

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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

#### SarahJeen

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

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

#### MarkFL

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

...
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 $$\displaystyle \tan(x)=0$$ 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.

#### SarahJeen

##### New member
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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Yes, 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.
2cosx = 1
cosx = .5

#### Prove It

##### Well-known member
MHB Math Helper
Re: Solve the following equations algebraically. Give solutions as exact values where 0≤x&lt;2π

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 \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*}?

#### SarahJeen

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

Yes, 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.
I put a new solution without the funny varibles

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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*}?
Yes ? am I correct

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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

#### Prove It

##### Well-known member
MHB Math Helper
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 \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*}?

#### MarkFL

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

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...