Solve Trig Equation: -sin(x)-2cos(2x)=0

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In summary, the problem is to solve for all x values between 0 and 2pi for the equation -sin(x)-2cos(2x)=0. The first step is to equate the equation to zero and find a way to factor it using identities for cos(2x). However, none of the identities seem to help and the equation cannot be factored easily. After some calculations, the solutions are found to be x = arcsin(1+sqrt(33)/8), x = arcsin(1-sqrt(33)/8), x = pi-arcsin(1+sqrt(33)/8), and x = pi-arcsin(1-sqrt(33)/8). However, it is
  • #1
verd
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Hey,

So... I have this problem:

-sin(x)-2cos(2x)

And I need to solve for all the x values between 0 and 2pi. Naturally, one would equate that to zero and get:

-sin(x)-2cos(2x)=0

And in order to find what that's equivalent to, you'd have to find a way to factor it... And in order to do that, you'd have to use identities... I know the following for cos(2x):

[tex]\cos(2x)=\cos(x)^2-\sin(x)^2[/tex]
[tex]\cos(2x)=2\cos(x)^2-1[/tex]
[tex]\cos(2x)=1-2\sin(x)^2[/tex]However, I don't see any of those actually helping. By using the last one, I get:

[tex]4\sin(x)^2-\sin(x)-2[/tex]

But there's no reasonable way to factor that. I went through the depths of complete the square hell to find this:

[tex]\frac{1}{2}(2\sin(x)-1)(4\sin(x)-1)-\frac{3}{2}=0[/tex]
[tex](2\sin(x)-1)(4\sin(x)-1)=3[/tex]

And if I set each parenthesis equal to 3, which I'm pretty sure you can't do, I get:

[tex]\arcsin(2)[/tex] and [tex]\arcsin(1/2)[/tex]

Both of which are wrong...Can anyone shed any light on this?
Haha, I'm having quite a bit of difficulty
 
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  • #2
verd said:
Hey,

So... I have this problem:

-sin(x)-2cos(2x)

And I need to solve for all the x values between 0 and 2pi.


Naturally, one would equate that to zero and get:

-sin(x)-2cos(2x)=0

And in order to find what that's equivalent to, you'd have to find a way to factor it... And in order to do that, you'd have to use identities... I know the following for cos(2x):

[tex]\cos(2x)=\cos(x)^2-\sin(x)^2[/tex]
[tex]\cos(2x)=2\cos(x)^2-1[/tex]
[tex]\cos(2x)=1-2\sin(x)^2[/tex]


However, I don't see any of those actually helping. By using the last one, I get:

[tex]4\sin(x)^2-\sin(x)-2[/tex]

But there's no reasonable way to factor that. I went through the depths of complete the square hell to find this:
It's fine till here, but how can't you factor that?
Do you know the formula to find the roots of a quadratic equation? Say, you want to solve for x in the equation:
ax2 + bx + c = 0.
[tex]x = \frac{-b \pm \sqrt{b ^ 2 - 4ac}}{2a}[/tex], does this formula look farmiliar to you? :)
Can you go from here? :)
 
  • #3
I tried that one first. That also didn't work. That gave me (1+sqrt33)/8, which you would put within the arcsin function

plugging it all in, it didn't work... instead of zero, i got 12 or something.
 
  • #4
verd said:
I tried that one first. That also didn't work. That gave me (1+sqrt33)/8, which you would put within the arcsin function

plugging it all in, it didn't work... instead of zero, i got 12 or something.
[tex]\frac{1 + \sqrt{33}}{8}[/tex] is correct.
So you'll have:
[tex]\left[ \begin{array}{l} \sin x = \frac{1 + \sqrt{33}}{8} \\ \sin x = \frac{1 - \sqrt{33}}{8} \end{array} \right.[/tex]
Can you find x? Note that [tex]x \in [0 , \ 2 \pi ][/tex] as the preblem states. There should be 4 x's. :)
 
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  • #5
right, wouldn't it be

[tex]\arcsin{x}=\frac{1+\sqrt(33)}{8}[/tex]

and viceaversa for the 1-sqrt33...however, when i plug that into the original equation, i don't get zero. this is what I'm plugging in:

[tex]-(\frac{1+\sqrt(33)}{8}-2\cos(\arcsin(\frac{1+\sqrt(33)}{8})))[/tex]

With an approximate value provided by the calculator, i get -1.9
 
  • #6
Ack, you've done it wrong...
It should be:
[tex]x = \arcsin \left( \frac{1 \pm \sqrt{33}}{8} \right)[/tex].
You should also note that:
[tex]\sin(x) = \sin (x + 2 \pi)[/tex]
[tex]\sin(x) = \sin (\pi - x)[/tex].
Can you find the 4 x's? :)
 
  • #7
I'm sorry... I still don't quite follow.

You're right on the x=arcsin(...) bit, I wrote that incorrectly... But I plugged the correct bit into the above equation...

I'm just a bit confused... It just looks like an impossible problem to solve, and I'm still stuck.
 
  • #8
I mean, I get your point, I get that it's

[tex]x = \arcsin \left( \frac{1 + \sqrt{33}}{8} \right)[/tex]
[tex]x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right)[/tex]

and then the other two have something to do with
[tex]x = \arcsin \left( \pi - \frac{1 + \sqrt{33}}{8} \right)[/tex]
[tex]x = \arcsin \left( \pi - \frac{1 - \sqrt{33}}{8} \right)[/tex]But when I went and plugged the first one into the original equation, I didn't get 0...

For cos[arcsin(...)]
Isn't there some sort of trigonometric trick or something? Or do I have to somehow use and label the triangle...?
 
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  • #9
But when I went and plugged the first one into the original equation, I didn't get 0...

Are you sure you're working in radians ? It won't make a difference in the first set of solutions, but it will make a difference in the second set (the ones with "pi" in them).

[tex]x = \arcsin \left( \frac{1 + \sqrt{33}}{8} \right)[/tex]
[tex]x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right)[/tex]

Those two are right.

and then the other two have something to do with

No, these should read

[tex]x = \pi - \arcsin \left(\frac{1 + \sqrt{33}}{8} \right)[/tex]
[tex]x = \pi - \arcsin \left(\frac{1 - \sqrt{33}}{8} \right)[/tex]
 
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  • #10
verd said:
I'm sorry... I still don't quite follow.

You're right on the x=arcsin(...) bit, I wrote that incorrectly... But I plugged the correct bit into the above equation...

I'm just a bit confused... It just looks like an impossible problem to solve, and I'm still stuck.
Uhmm, where did you get stuck?
Okey, I'll give you an example:
Find [tex]x \in [0 , \ \pi ][/tex], such that sin(x) = 0.4
All the x that satisfy the above equation is:
[tex]\left[ \begin{array}{l} x = \arcsin (0.4) + 2k \pi = 0.411 + 2k \pi \\ x = \pi - \arcsin (0.4) + 2m \pi = 2.73 + 2m \pi \end{array} \right. k, \ m \in \mathbb{Z}[/tex]
Plug some value of k into see if [tex]x \in [0 , \ \pi ][/tex].
k = -1 ----> x = -5.8
k = 0 ----> x = 0.41
k = 1 ----> x = 6.7
So k = 0 will give 1 valid x.
Do the same for m, and you will have another valid x: x = 2.7 when m = 0.
So the 2 solutions are:
[tex]\left[ \begin{array}{l} x = \arcsin (0.4) \\ x = \pi - \arcsin (0.4) \end{array} \right.[/tex]
Can you get this? :)
------------------
Note that:
[tex]x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right) < 0[/tex]
By the way, what do you get if you plug that x in?
Remember that it is:
-sin x - 2cos(2x)
Don't forget the factor of 2. :)
 
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  • #11
You're right, I was missing the 2 infront of the x within the cosine function...

However, now I get a really small number. The calculator is returning -1.9e-13

I mean, it's close to zero... i mean, is it close enough?
 
  • #12
That's ok, it's just rounding error, don't worry about it.
Have you found out the 4 x's? :)
 
  • #13
arcsin(1+...)
arcsin(1-...)

pi-arcsin(1+...)
pi-arcsin(1-...)Correct?Thanks for all of your help!
 
  • #14
As I said earlier (in post # 10) that:
[tex]x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right) < 0[/tex] is not a valid solution by the restriction [tex]x \in [ 0, \ 2 \pi ][/tex]
Remember that sine is periodic function with period [tex]2 \pi[/tex], you can go from here? Right? :)
 
  • #15
well sure, I guess it would then be:

pi-arcsin(1+...)
arcsin(1+...)

2pi+arcsin(1-...)
pi-arcsin(1-...)

would this be correct?
all are greater than 0, i believe...
 
  • #16
That's correct. Congratulations. :)
 

1. How do I solve a trigonometric equation?

To solve a trigonometric equation like -sin(x)-2cos(2x)=0, you need to use algebraic techniques to isolate the variable on one side of the equation. Then, you can use inverse trigonometric functions or trigonometric identities to solve for the variable.

2. What is the first step in solving this equation?

The first step in solving -sin(x)-2cos(2x)=0 is to distribute the negative sign to the terms inside the parentheses, giving you -sin(x)-2cos(x)cos(x)=0.

3. Can I use a calculator to solve this equation?

While you can use a calculator to check your work, it is not recommended to use a calculator to solve -sin(x)-2cos(2x)=0. This is because the solution involves finding the inverse trigonometric functions of a number, which can be difficult for a calculator to accurately display.

4. Are there any restrictions on the values of x for this equation?

Yes, there are restrictions on the values of x for -sin(x)-2cos(2x)=0. Since the cosine function has a range of -1 to 1, the value of -2cos(2x) can only be between -2 and 2. This means that the values of x must make the expression -sin(x)-2cos(2x) equal to 0 within this range.

5. Can I use the quadratic formula to solve this equation?

Yes, you can use the quadratic formula to solve -sin(x)-2cos(2x)=0. After distributing the negative sign and rearranging the terms, you will have a quadratic equation in the form of ax^2+bx+c=0, where a=1, b=-2cos(x), and c=-sin(x). You can then use the quadratic formula x = (-b ± √(b^2-4ac)) / 2a to find the solutions for x.

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