Find the general solution for...

**Educated** · November 4th 2011, 07:34 PM

I'm having a bit of trouble solving this trig problem:

Find the General solution for... $\sin(4x) - \cos(x) = 0$

At first when I looked at it it seemed easy but as I tried to solve it, it got really confusing and I got stuck on the $\sin^3(x)$ part.

Here's what I did:
Double angle formula: $\sin(2 \theta) = 2 \sin (\theta) \cos (\theta)$
Double angle formula: $\cos(2 \theta) = 1-2\sin^2(\theta)$

$0=\sin(4x) - \cos(x)$
$0=2\cdot\sin(2x)\cdot\cos(2x) - \cos(x)$
$0=4[sin(x)\cos(x)]\cdot[1-\sin^2(x)] - \cos(x)$
$0=4\sin(x)\cos(x) - 8\sin^3(x)\cos(x) - \cos(x)$

$0=\cos(x)[4\sin(x) - 8\sin^3(x) - 1]$

Now I can get the general solution for the $\cos(x)$ part but what about the $4\sin(x) - 8\sin^3(x) - 1$ part? How would I solve that?

EDIT: Sorry I just realised I placed this in the wrong section :/
Can a mod please move it to the trig section?

**Prove It** · November 4th 2011, 07:48 PM

Originally Posted by Educated

I'm having a bit of trouble solving this trig problem:

At first when I looked at it it seemed easy but as I tried to solve it, it got really confusing and I got stuck on the $\sin^3(x)$ part.

Here's what I did:
Double angle formula: $\sin(2 \theta) = 2 \sin (\theta) \cos (\theta)$
Double angle formula: $\cos(2 \theta) = 1-2\sin^2(\theta)$

$0=\sin(4x) - \cos(x)$
$0=2\cdot\sin(2x)\cdot\cos(2x) - \cos(x)$
$0=4[sin(x)\cos(x)]\cdot[1-\sin^2(x)] - \cos(x)$
$0=4\sin(x)\cos(x) - 8\sin^3(x)\cos(x) - \cos(x)$

$0=\cos(x)[4\sin(x) - 8\sin^3(x) - 1]$

Now I can get the general solution for the $\cos(x)$ part but what about the $4\sin(x) - 8\sin^3(x) - 1$ part? How would I solve that?

EDIT: Sorry I just realised I placed this in the wrong section :/
Can a mod please move it to the trig section?

You won't be able to solve $\displaystyle 8\sin^3{x} - 4\sin{x} + 1 = 0$ using analytical methods. You will need to use technology.

**BAdhi** · November 4th 2011, 10:29 PM

Actually, $\sin{x}=\frac{1}{2}$ seems to satisfy $8\sin^3{x}-4\sin{x}+1=0$ .

**Prove It** · November 4th 2011, 10:42 PM

Originally Posted by BAdhi

Actually, $\sin{x}=\frac{1}{2}$ seems to satisfy $8\sin^3{x}-4\sin{x}+1=0$ .

Yes, that's true. However, if you are not a very good guesser, technology would be the way to go

**Educated** · November 5th 2011, 12:49 AM

This was an old exam question so we're not allowed to use technology to solve it, just our brains.

Although I guess you can guess one solution by trial and error for the cubic $8u^3 -4u +1$ where $u=\sin(x)$ and try to derive the other solutions from it by factoring.

Thanks for the help!

**sbhatnagar** · November 5th 2011, 03:22 AM

$\\8u^3-4u+1=0 \\ (2u-1)(4u^2+2u-1)=0\\u=\frac{1}{2},\hspace{10}\frac{1}{4}(\pm \sqrt{5} -1)$

Therefore, $\sin{x}= \frac{1}{2},\hspace{10}\frac{1}{4}(\sqrt{5}-1),\hspace{10}\frac{1}{4}(-\sqrt{5}-1)$ .

$x=\arcsin{[\frac{1}{2}]}+2k\pi$

$x=\arcsin{[\frac{1}{4}(\sqrt{5}-1)]}+2k\pi$

$x=\arcsin{[\frac{1}{4}(-\sqrt{5}-1)]}+2k\pi$

where k is some integer.

**Soroban** · November 5th 2011, 06:00 AM

Hello, Educated!

$\text{Find the general solution for: }\:\sin(4x) - \cos(x) \:=\:0$

Here's a little help . . .

Since $\sin(4x) \,=\,\cos(x)$ , then $4x$ and $x$ are in the same right triangle.

Code:

Hence: . $4x + x \,=\,\tfrac{\pi}{2} \quad\Rightarrow\quad 5x \,=\,\tfrac{\pi}{2} \quad\Rightarrow\quad x \,=\,\tfrac{\pi}{10}$

I know . . . I oversimplified the problem.
. . But you can generalize the solution, right?

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