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Trigonometry problem. Solve equation for

  • Thread starter SolCon
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  • #1
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hello again. :smile:

I have 2 trigno "solve the equation for" problems which I am having difficulty with.

Q.1) This came in 2 parts. I've done the first part but I'll post it anyway as it is important for the next part (where I'm having difficulty).


i) Show that equation 4sin^4[tex]\theta[/tex] + 5=7cos^2[tex]\theta[/tex] may be written as 4x^2+7x-2=0 where x=sin^2[tex]\theta[/tex].

I got this part done.

ii) then it says to solve the equation 4sin^4[tex]\theta[/tex] + 5=7cos^2[tex]\theta[/tex] for 0 <= [tex]\theta[/tex] <= 360.

I tried factorising the proven equation (with the x only) and received 2 values of x i.e x=1/4
and x=-2. The x=-2 is undefined but for the x=1/4, this seems to be incorrect but I do not know why.

Q.2) This is similar to part II of the first question. It says to find all the values of x in the interval 0<=x<=180 which satisfies the equation sin3x+2cos3x=0.

For this one, I simply shifted 2cos3x to the other side and simplified it as tan3x=-2. Is this correct? If it is, I'm still getting wrong values. Do we take this as x=tan^-1(-2/3) or some other way? If the former, then the value I get is -33.69 which seems to be incorrect (correct base value is 38.9).

Any help with both these questions? :confused:
 

Answers and Replies

  • #2
Char. Limit
Gold Member
1,204
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You say that x=1/4 is "incorrect". However, are you sure? You get an equation sin(x)=1/2 or sin(x)=-1/2.

And for the second one, take the inverse tangent and then divide by 3.
 
  • #3
33
1
Thanks for the reply. :smile:

however, I'm still confused.

1) the equation will be Sinx=1/4. Therefore x=sin^-1(1/4). This comes as 14.47 which is not the correct base value. You have said Sinx=1/2. This gives us 30 which is the correct value. But how did you get it? I'll show you what I did:

4x^2+7x-2=0 [we'll factorise this.]
4x^2+8x-x-2=0
4x(x+2) -1(x+2)=0

hence, 4x-1=0 and x+2=0
x=1/4 and x=-2

apply Sin() for both

Sinx=1/4 ; Sinx=-2
x=Sin^-1(1/4) ; x=Sin^-1(-2)
x=14.47 ; x=Undefined.

This is how I'm doing it. Could you show me how you got Sinx=1/2?

2) I'm still not getting the right answer.

tan3x=-2
3x=tan^-1(-2) [like you have said]
3x=-63.4
x=-21.14

However, the correct answer is 38.9. So what could be wrong here?
 
  • #4
eumyang
Homework Helper
1,347
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Thanks for the reply. :smile:

however, I'm still confused.

1) the equation will be Sinx=1/4. Therefore x=sin^-1(1/4). This comes as 14.47 which is not the correct base value. You have said Sinx=1/2. This gives us 30 which is the correct value. But how did you get it? I'll show you what I did:

4x^2+7x-2=0 [we'll factorise this.]
4x^2+8x-x-2=0
4x(x+2) -1(x+2)=0

hence, 4x-1=0 and x+2=0
x=1/4 and x=-2

apply Sin() for both

Sinx=1/4 ; Sinx=-2
Here's the problem. Substitute in sin2θ for x, not sin θ. Then take the square-root of both sides and apply the inverse sine.

tan3x=-2
3x=tan^-1(-2) [like you have said]
3x=-63.4
x=-21.14
The problem is keeping 3x = -63.4° (I'm assuming degrees here). Remember that 0° ≤ x < 180°. So 0° ≤ 3x < 540° (multiply everything by 3). But you have 3x = -63.4°, which is not in the accepted range. Add a number to this to fix it. (Hint: what is the period of tan x?)
 
Last edited:
  • #5
33
1
Can't believe I missed that... :redface:

As for the second part.

I believe the period of tan(x) is pi. So.... pi/3 or 3pi will be used in some way?
 
  • #6
VietDao29
Homework Helper
1,423
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Can't believe I missed that... :redface:

As for the second part.

I believe the period of tan(x) is pi. So.... pi/3 or 3pi will be used in some way?
Yup, the period of tan(x) is pi. Which means that:
[tex]\tan (\alpha) = \tan(\alpha + \pi) = ... = \tan (\alpha + k \pi), \forall k \in \mathbb{Z}[/tex] (1)

So, the general solution of the following equation:
[tex]\tan(x) = \rho \Leftrightarrow x = \arctan (\rho) + k \pi[/tex]
This stems from the fact that [tex]\arctan (\rho)[/tex] is one solution to the equation, and [tex]\tan(\alpha) = \tan(\alpha + k \pi)[/tex], for all integer k (see (1)).

So, can you do the same to find the general solution to the equation:
tan(3x) = -2? And then choose the appropriate values for k, such that your solution is in the required interval?
 
  • #7
33
1
Yup, the period of tan(x) is pi. Which means that:
[tex]\tan (\alpha) = \tan(\alpha + \pi) = ... = \tan (\alpha + k \pi), \forall k \in \mathbb{Z}[/tex] (1)

So, the general solution of the following equation:
[tex]\tan(x) = \rho \Leftrightarrow x = \arctan (\rho) + k \pi[/tex]
This stems from the fact that [tex]\arctan (\rho)[/tex] is one solution to the equation, and [tex]\tan(\alpha) = \tan(\alpha + k \pi)[/tex], for all integer k (see (1)).

So, can you do the same to find the general solution to the equation:
tan(3x) = -2? And then choose the appropriate values for k, such that your solution is in the required interval?
You may find this annoying but...

All I can get is 3x=tan^-1(-2). No idea what to do next or even if this is correct.
 
  • #8
eumyang
Homework Helper
1,347
10
Consider this:
If tan x = 1, then x is not only 45°, 45° plus an integer multiple of 180°, because the period of tangent is 180° (pi). So x = ..., -315°, -135°, 45°, 225°, 405°...

You earlier had
3x = tan^-1(-2)
3x = -63.4°,
and this wouldn't be wrong, if not for the restriction given for x (and 3x). Remember that 0° ≤ 3x < 540°. So take -63.4° and add the period for tangent to it to put 3x within the accepted range. Then divide both sides by 3.
 
  • #9
VietDao29
Homework Helper
1,423
2
You may find this annoying but...

All I can get is 3x=tan^-1(-2). No idea what to do next or even if this is correct.
Since the problem is asking for x, such that 00 <= x <= 1800, I'll use degrees here.

We know that: [tex]3x = \arctan(-2)[/tex] is one solution to the equation tan(3x) = -2. And since we have:

[tex]\tan(\alpha) = \tan(\alpha + k 180 ^ {\circ}), \forall k \in \mathbb{Z}[/tex]. This is because the period for tan(x) is 1800.

Which means that: [tex]\tan(3x) = \tan(3x + k 180 ^ {\circ}), \forall k \in \mathbb{Z}[/tex].

So [tex]3x = \arctan(-2) + k.180^{\circ}, \forall k \in \mathbb{Z}[/tex] is also a solution to the equation: tan(3x) = -2.

[tex]3x = \arctan(-2) + k.180^{\circ} \Leftrightarrow x = \frac{\arctan(-2)}{3} + k.60^{\circ}[/tex]

Now, can you find the possible velues for k such that 00 <= x <= 1800?

I hope that everything is clear now. If you find somewhere unclear, just ask, don't be shy. :)
 
  • #10
33
1
Apologies for the late response but I had some things to take care of. :cry:

Anyways;

eumyang:

That solution works. What you said:

3x=tan^-1(-2)
3x=-63.43

As tan's period is pi, we add 180+ constantly to the value till we reach the maximum under 540 (step 1 below) and divide those by 3 (step 2 below). So;

1) -63.43+180 = 116
116+180 = 296
296+180=476 (Stopping here as next value is greater than 540.
476+180=656

and so on..

2) 116/3= 38.66 (close enough :smile: )
296/3= 98.6
476/3= 158.6

And these are the values listed as the correct answer.

However, I have to ask. Why did we not divide -63.43 by 3. Sometimes, we do divide the value do we not? Like in this question:

Solve equation cos2x -= -0.7 for x in the interval 0 <=x<=360. The first value here is 67.2 which we get by:

cos2x=-0.7
2x= cos^-1(-0.7)
2x=134.4
x=134.4/2
x=67.2

This is the correct value as the question states so. So how is it that we had to divide here but not above in the case of tan3x=-2

VietDao29:

I'm still stuck. :confused:

What I can get is:

3x=tan^-1(-2)+k.180 [like you have stated]
x=[tan^-1(-2)]/3 + k.60 [again, like you have done above]
x=-63.43+k.60

What to do next. It isn't quad, so b2-4ac can't come here. There are 2 unknowns. Is k, by any chance, 1? Since period of tan is pi or 1(pi) and you have said: tan(a+k.pi)??
 
  • #11
Char. Limit
Gold Member
1,204
14
However, I have to ask. Why did we not divide -63.43 by 3. Sometimes, we do divide the value do we not? Like in this question:

Solve equation cos2x -= -0.7 for x in the interval 0 <=x<=360. The first value here is 67.2 which we get by:

cos2x=-0.7
2x= cos^-1(-0.7)
2x=134.4
x=134.4/2
x=67.2

This is the correct value as the question states so. So how is it that we had to divide here but not above in the case of tan3x=-2
The reason you didn't use -63.43 as an accepted value here is because it didn't fall within the range of 0<3x<540. However, the other three values you used did. Now, it's different with the cosine function because cos-1(-.7) HAS values within the range of 0<x<360.

VietDao29:

I'm still stuck. :confused:

What I can get is:

3x=tan^-1(-2)+k.180 [like you have stated]
x=[tan^-1(-2)]/3 + k.60 [again, like you have done above]
x=-63.43+k.60

What to do next. It isn't quad, so b2-4ac can't come here. There are 2 unknowns. Is k, by any chance, 1? Since period of tan is pi or 1(pi) and you have said: tan(a+k.pi)??
k is any integer. So just plug in some integer values for k. Try 0, 1, 2, and 3.
 
  • #12
33
1
The reason you didn't use -63.43 as an accepted value here is because it didn't fall within the range of 0<3x<540. However, the other three values you used did. Now, it's different with the cosine function because cos-1(-.7) HAS values within the range of 0<x<360.
So in the case of tan3x, we didn't divide -63.43 because it was a negative value (and seeing as how a -ve value cannot fit into 0<=x<=180 or in the later case, 0<3x<540)?

And in the case of cos2x, we could divide 67.2 because it was a positive value?

Therefore, is the rule such that if there is a negative base value for bx= Sin, Cos or Tan inverse(the other value here), we multiply the higher range limit with 'b' (the coefficient of x like in tan3x or cos2x etc) and plug the bx value in the middle then follow the procedure I have done in my previous post? So in the case of cos2x, if it was a -ve value, we would have multiplied the higher range limit (360) by 2 and received 720 then followed the same procedure as done with tan3x in the question?

Is that right?



k is any integer. So just plug in some integer values for k. Try 0, 1, 2, and 3.
Right, but which value(s) should we take and why? And if we take 0,1,2, and 3 we get for these:

0: x=-63.43
1: x=-3.43
2: x=56.53
3: x=116.53 [got this above]
6: x=296.53 [got this above]
9: x=476.53 [got this above]

Okay, wait. These values are what I got above. So we have to take the multiples of 3 (3,6 and 9) because of tan3x? Then divide them by 3?
 
  • #13
eumyang
Homework Helper
1,347
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Right, but which value(s) should we take and why? And if we take 0,1,2, and 3 we get for these:

0: x=-63.43
1: x=-3.43
2: x=56.53
3: x=116.53 [got this above]
6: x=296.53 [got this above]
9: x=476.53 [got this above]

Okay, wait. These values are what I got above. So we have to take the multiples of 3 (3,6 and 9) because of tan3x? Then divide them by 3?
Careful, you're mixing x with 3x. Going back to post #10:

VietDao29:

I'm still stuck. :confused:

What I can get is:

3x=tan^-1(-2)+k.180 [like you have stated]
x=[tan^-1(-2)]/3 + k.60 [again, like you have done above]
x=-63.43+k.60

What to do next. It isn't quad, so b2-4ac can't come here. There are 2 unknowns. Is k, by any chance, 1? Since period of tan is pi or 1(pi) and you have said: tan(a+k.pi)??
The bolded is wrong. You forgot to divide by 3. So
[tex]x \approx -21.14^{\circ} + k \cdot 60^{\circ}[/tex]
Now if you plug in 1, 2, and 3 for k you get the correct answers for x:
x ≈ 38.86°, 98.86°, 158.86°
 
  • #14
33
1
Alright, getting close to finishing this. Just 2 things:

1) In the method posted by VietDao29, do we apply it for tan only? What would a similar formula be for sin and cos.

2) I'm still confused about the range multiplication bit. In this tan3x question, the range was 0<=x<=180. We have multiplied the the values of x and 180 with 3 because of tan3x. So it then becomes 0<3x<540. However, in the case of the example with cos2x=-0.7, the range was 0<=x<=360. Yet, we did not multiply 360 by 2 to get 720 and have 2x in the middle. Why is this so?
 
  • #15
eumyang
Homework Helper
1,347
10
Alright, getting close to finishing this. Just 2 things:

1) In the method posted by VietDao29, do we apply it for tan only? What would a similar formula be for sin and cos.

2) I'm still confused about the range multiplication bit. In this tan3x question, the range was 0<=x<=180. We have multiplied the the values of x and 180 with 3 because of tan3x. So it then becomes 0<3x<540. However, in the case of the example with cos2x=-0.7, the range was 0<=x<=360. Yet, we did not multiply 360 by 2 to get 720 and have 2x in the middle. Why is this so?
It is NOT so. Dealing with sine and cosine is a little more tricky. I overlooked that problem, and actually, you are missing some solutions.

Solve equation cos2x -= -0.7 for x in the interval 0 <=x<=360. The first value here is 67.2 which we get by:

cos2x=-0.7
2x= cos^-1(-0.7)
2x=134.4
x=134.4/2
x=67.2

This is the correct value as the question states so. So how is it that we had to divide here but not above in the case of tan3x=-2
The period for cosine is 2pi. Since 0° ≤ x < 360°, 0° ≤ 2x < 720°. So you'll have to add 360° to 134.4° to get a 2nd answer.
2x ≈ 134.4° OR
2x ≈ 134.4° + 360° ≈ 494.4°

But wait, there's more! Cosine is negative in the 2nd and 3rd quadrant. 134.4° is in the 2nd quadrant, and it's related angle is 180° - 134.4° ≈ 45.6°. You need to find the 3rd quadrant angle whose related angle is 45.6°. It's 180 + 45.6° ≈ 225.6°. So that's the 3rd answer. And finally, add 360° to that to get the fourth answer.

cos 2x = -0.7
2x = cos^-1 (-0.7)

2x ≈ 134.4°
OR 2x ≈ 225.6°
OR 2x ≈ 134.4° + 360° ≈ 494.4°
OR 2x ≈ 225.6° + 360° ≈ 585.6°

Divide all of these angles by 2 to get your four solutions for x.
 
  • #16
33
1
The values we get are:
134.4/2,225.6/2,494.4/2,585.6/2
=67.2,112.5,247.2,292.6

...all of which are correct.

So, what I finally understand is this:

For those cases where 'b' in 'bx' is greater than one in sin or cos (2x, 3x etc), we apply the method you've said, where we don't divide the inverse value with b but instead multiply it out with the maximum of the range and with the x in the middle. Then we use the 2nd, 3rd, 4th quadrant method (180-,180+,360+) to get respective values till we reach just under the maximum then divide those values by b.

And for tan, we get the value then keep adding 180 to them till we reach just under the maximum then divide those values by b.

Is that right?
 

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