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

[verd]

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):

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

$$4\sin(x)^2-\sin(x)-2$$

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

$$\frac{1}{2}(2\sin(x)-1)(4\sin(x)-1)-\frac{3}{2}=0$$
$$(2\sin(x)-1)(4\sin(x)-1)=3$$

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

$$\arcsin(2)$$ and $$\arcsin(1/2)$$

Both of which are wrong...Can anyone shed any light on this?
Haha, I'm having quite a bit of difficulty

 

[VietDao29]

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):

$$\cos(2x)=\cos(x)^2-\sin(x)^2$$
$$\cos(2x)=2\cos(x)^2-1$$
$$\cos(2x)=1-2\sin(x)^2$$


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

$$4\sin(x)^2-\sin(x)-2$$

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.
$$x = \frac{-b \pm \sqrt{b ^ 2 - 4ac}}{2a}$$, does this formula look farmiliar to you? :)
Can you go from here? :)

 

[verd]

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.

 

[VietDao29]

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.

$$\frac{1 + \sqrt{33}}{8}$$ is correct.
So you'll have:
$$\left[ \begin{array}{l} \sin x = \frac{1 + \sqrt{33}}{8} \\ \sin x = \frac{1 - \sqrt{33}}{8} \end{array} \right.$$
Can you find x? Note that $$x \in [0 , \ 2 \pi ]$$ as the preblem states. There should be 4 x's. :)

 

[verd]

right, wouldn't it be

$$\arcsin{x}=\frac{1+\sqrt(33)}{8}$$

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:

$$-(\frac{1+\sqrt(33)}{8}-2\cos(\arcsin(\frac{1+\sqrt(33)}{8})))$$

With an approximate value provided by the calculator, i get -1.9

 

[VietDao29]

Ack, you've done it wrong...
It should be:
$$x = \arcsin \left( \frac{1 \pm \sqrt{33}}{8} \right)$$.
You should also note that:
$$\sin(x) = \sin (x + 2 \pi)$$
$$\sin(x) = \sin (\pi - x)$$.
Can you find the 4 x's? :)

 

[verd]

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.

 

[verd]

I mean, I get your point, I get that it's

$$x = \arcsin \left( \frac{1 + \sqrt{33}}{8} \right)$$
$$x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right)$$

and then the other two have something to do with
$$x = \arcsin \left( \pi - \frac{1 + \sqrt{33}}{8} \right)$$
$$x = \arcsin \left( \pi - \frac{1 - \sqrt{33}}{8} \right)$$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...?

 

[Curious3141]

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).

$$x = \arcsin \left( \frac{1 + \sqrt{33}}{8} \right)$$
$$x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right)$$

Those two are right.

and then the other two have something to do with

No, these should read

$$x = \pi - \arcsin \left(\frac{1 + \sqrt{33}}{8} \right)$$
$$x = \pi - \arcsin \left(\frac{1 - \sqrt{33}}{8} \right)$$

 

[VietDao29]

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 $$x \in [0 , \ \pi ]$$, such that sin(x) = 0.4
All the x that satisfy the above equation is:
$$\left[ \begin{array}{l} x = \arcsin (0.4) + 2k \pi = 0.411 + 2k \pi \\ x = \pi - \arcsin (0.4) + 2m \pi = 2.73 + 2m \pi \end{array} \right. k, \ m \in \mathbb{Z}$$
Plug some value of k into see if $$x \in [0 , \ \pi ]$$.
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:
$$\left[ \begin{array}{l} x = \arcsin (0.4) \\ x = \pi - \arcsin (0.4) \end{array} \right.$$
Can you get this? :)
------------------
Note that:
$$x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right) < 0$$
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. :)

 

[verd]

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?

 

[VietDao29]

That's ok, it's just rounding error, don't worry about it.
Have you found out the 4 x's? :)

 

[verd]

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

pi-arcsin(1+...)
pi-arcsin(1-...)Correct?Thanks for all of your help!

 

[VietDao29]

As I said earlier (in post # 10) that:
$$x = \arcsin \left( \frac{1 - \sqrt{33}}{8} \right) < 0$$ is not a valid solution by the restriction $$x \in [ 0, \ 2 \pi ]$$
Remember that sine is periodic function with period $$2 \pi$$, you can go from here? Right? :)

 

[verd]

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...

 

[VietDao29]

That's correct. Congratulations. :)