Find the general solution for...

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

Quote:

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

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 $\displaystyle \sin^3(x)$ part.

Here's what I did:

Double angle formula: $\displaystyle \sin(2 \theta) = 2 \sin (\theta) \cos (\theta)$

Double angle formula: $\displaystyle \cos(2 \theta) = 1-2\sin^2(\theta)$

$\displaystyle 0=\sin(4x) - \cos(x)$

$\displaystyle 0=2\cdot\sin(2x)\cdot\cos(2x) - \cos(x)$

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

$\displaystyle 0=4\sin(x)\cos(x) - 8\sin^3(x)\cos(x) - \cos(x)$

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

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

Re: Find the general solution for...

Quote:

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 $\displaystyle \sin^3(x)$ part.

Here's what I did:

Double angle formula: $\displaystyle \sin(2 \theta) = 2 \sin (\theta) \cos (\theta)$

Double angle formula: $\displaystyle \cos(2 \theta) = 1-2\sin^2(\theta)$

$\displaystyle 0=\sin(4x) - \cos(x)$

$\displaystyle 0=2\cdot\sin(2x)\cdot\cos(2x) - \cos(x)$

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

$\displaystyle 0=4\sin(x)\cos(x) - 8\sin^3(x)\cos(x) - \cos(x)$

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

Now I can get the general solution for the $\displaystyle \cos(x)$ part but what about the $\displaystyle 4\sin(x) - 8\sin^3(x) - 1$ 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 $\displaystyle \displaystyle 8\sin^3{x} - 4\sin{x} + 1 = 0$ using analytical methods. You will need to use technology.

Re: Find the general solution for...

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

Re: Find the general solution for...

Quote:

Originally Posted by

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

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

Re: Find the general solution for...

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 $\displaystyle 8u^3 -4u +1$ where $\displaystyle u=\sin(x)$ and try to derive the other solutions from it by factoring.

Thanks for the help!

Re: Find the general solution for...

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

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

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

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

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

where k is some integer.

Re: Find the general solution for...

Hello, Educated!

Quote:

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

Here's a little help . . .

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

Code:

` *`

/|

/x|

/ |

/ |

/ |

/ |

/4x |

* - - - *

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

I know . . . I oversimplified the problem.

. . But you can generalize the solution, right?