Trig equation: sin(5x)+sin(x)=0

Question: Solve $\displaystyle \displaystyle sin(5x)+sin(x)=0$ for $\displaystyle 0\leq x < \pi$.

I know ways to solve this by expanding sin(5x) but I'm wondering if the method I'm attempting below is correct and how to proceed with it:

$\displaystyle \displaystyle \sin 5x = -sin x \implies \sin 5x = \sin {-x}$

$\displaystyle \implies \sin 5(x-2k_1 \pi) = \sin -(x-2k_2 \pi)$

where $\displaystyle k_1$ and $\displaystyle k_2$ are integers.

$\displaystyle \displaystyle \implies 5x-10k_1 \pi = -x+2k_2 \pi \implies x=\frac{\pi(2k_2+10k_1)}{6} = \frac{\pi(k_2+5k_1)}{3}$

Now the solution interval is between 0 and pi so we have the following inequality:

$\displaystyle \displaystyle 0\leq \frac{k_2+5k_1}{3}<1 \implies 0\leq k_2+5k_1<3$

There are infinite integer solutions to this inequality so how do I limit the solutions further?

Re: Trig equation: sin(5x)+sin(x)=0

Quote:

Originally Posted by

**Josh146** Question: Solve $\displaystyle \displaystyle sin(5x)+sin(x)=0$ for $\displaystyle 0\leq x < \pi$.

I know ways to solve this by expanding sin(5x) but I'm wondering if the method I'm attempting below is correct and how to proceed with it:

$\displaystyle \displaystyle \sin 5x = -sin x \implies \sin 5x = \sin {-x}$ **<--- hmmmm, I never heard about that identity**

...

Here is how I would have done this question:

1. $\displaystyle \sin(5x)= 16\sin(x) \cdot (\cos(x))^4 - 12\sin(x) \cdot (\cos(x))^2+\sin(x)$

2. That means your equation becomes:

$\displaystyle 16\sin(x) \cdot (\cos(x))^4 - 12\sin(x) \cdot (\cos(x))^2+2\sin(x)=0$

3. Factor out 2sin(x):

$\displaystyle 2\sin(x)\left(8 \cdot (\cos(x))^4 - 6 \cdot (\cos(x))^2+1 \right)=0$

4. A product equals zero if one of the factors equals zero:

$\displaystyle 2\sin(x) = 0~\vee~\left(8 \cdot (\cos(x))^4 - 6 \cdot (\cos(x))^2+1 \right)=0$

5. For the 2nd equation use the substitution $\displaystyle u = (\cos(x))^2$ . So you have to solve for u:

$\displaystyle 8u^2-6u+1=0$

which yields 2 positive solutions. Use them to determine x.

Re: Trig equation: sin(5x)+sin(x)=0

Quote:

Originally Posted by

**Josh146** Question: Solve $\displaystyle \displaystyle sin(5x)+sin(x)=0$ for $\displaystyle 0\leq x < \pi$.

I know ways to solve this by expanding sin(5x) but I'm wondering if the method I'm attempting below is correct and how to proceed with it:

$\displaystyle \displaystyle \sin 5x = -sin x \implies \sin 5x = \sin {-x}$

$\displaystyle \implies \sin 5(x-2k_1 \pi) = \sin -(x-2k_2 \pi)$

where $\displaystyle k_1$ and $\displaystyle k_2$ are integers.

$\displaystyle \displaystyle \implies 5x-10k_1 \pi = -x+2k_2 \pi \implies x=\frac{\pi(2k_2+10k_1)}{6} = \frac{\pi(k_2+5k_1)}{3}$

Now the solution interval is between 0 and pi so we have the following inequality:

$\displaystyle \displaystyle 0\leq \frac{k_2+5k_1}{3}<1 \implies 0\leq k_2+5k_1<3$

There are infinite integer solutions to this inequality so how do I limit the solutions further?

Here's how I would do it... Using $\displaystyle \displaystyle \begin{align*} \sin{\left(n\theta\right)} &= \sum_{k = 0}^n{{n\choose{k}}\cos^k{(\theta)}\sin^{n-k}{(\theta)}\sin{\left[\frac{1}{2}(n - k)\pi \right]}} \end{align*} $ we have

$\displaystyle \displaystyle \begin{align*} \sin{(5\theta)} &= \sum_{k = 0}^5{{5\choose{k}}\cos^k{(\theta)}\sin^{5-k}{(\theta)}\sin{\left[\frac{1}{2}(5-k)\pi\right]}} \\ &= \sin^5{(\theta)}\sin{ \left( \frac{5\pi}{2} \right) } + 5\cos{(\theta)}\sin^4{(\theta)}\sin{ \left( 2\pi \right) } + 10\cos^2{(\theta)}\sin^3{(\theta)}\sin{ \left( \frac{3\pi}{2} \right) } + 10\cos^3{(\theta)}\sin^2{(\theta)}\sin{\left(\pi \right)} + 5\cos^4{(\theta)}\sin{(\theta)}\sin{ \left( \frac{\pi}{2} \right) } + \cos^5{(\theta)}\sin{(0)} \\ &= \sin^5{(\theta)} - 10\cos^2{(\theta)}\sin^3{(\theta)} + 5\cos^4{(\theta)}\sin{(\theta)} \end{align*} $

Therefore

$\displaystyle \displaystyle \begin{align*} \sin{(5x)} + \sin{(x)}&= 0 \\ \sin^5{(x)} - 10\cos^2{(x)}\sin^3{(x)} + 5\cos^4{(x)}\sin{(x)} + \sin{(x)} &= 0 \\ \sin{(x)}\left[\sin^4{(x)} - 10\cos^2{(x)}\sin^2{(x)} + 5\cos^4{(x)} + 1\right] &= 0 \\ \sin{(x)}\left\{ \sin^4{(x)} - 10\left[ 1 - \sin^2{(x)} \right]\sin^2{(x)} + 5\left[ 1 - \sin^2{(x)} \right]^2 + 1 \right\} &= 0 \\ \sin{(x)}\left[ \sin^4{(x)} - 10\sin^2{(x)} + 10\sin^4{(x)} + 5 - 10\sin^2{(x)} + 5\sin^4{(x)} + 1 \right] &= 0 \\ \sin{(x)}\left[ 16\sin^4{(x)} - 20\sin^2{(x)} + 6 \right] &= 0 \\ 2\sin{(x)}\left[8\sin^4{(x)} - 10\sin^2{(x)} + 3\right] &= 0 \\ 2\sin{(x)}\left[ 8\sin^4{(x)} - 6\sin^2{(x)} - 4\sin^2{(x)} + 3 \right] &= 0 \\ 2\sin{(x)}\left\{ 2\sin^2{(x)}\left[ 4\sin^2{(x)} - 3 \right] - 1\left[ 4\sin^2{(x)} - 3 \right] \right\} &= 0 \\ 2\sin{(x)}\left[ 4\sin^2{(x)} - 3 \right]\left[ 2\sin^2{(x)} - 1 \right] &= 0\end{align*} $

Case 1:

$\displaystyle \displaystyle \begin{align*} \sin{(x)} &= 0 \\ x &= \left\{ 0, \pi \right\} + 2\pi n, n \in \mathbf{Z} \end{align*} $

Case 2:

$\displaystyle \displaystyle \begin{align*} 4\sin^2{(x)} - 3 &= 0 \\ 4\sin^2{(x)} &= 3 \\ \sin^2{(x)} &= \frac{3}{4} \\ \sin{(x)} &= \pm \frac{\sqrt{3}}{2} \\ x &= \left\{ \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \right\} + 2\pi n, n \in \mathbf{Z} \end{align*} $

Case 3:

$\displaystyle \displaystyle \begin{align*} 2\sin^2{(x)} - 1 &= 0 \\ 2\sin^2{(x)} &= 1 \\ \sin^2{(x)} &= \frac{1}{2} \\ \sin{(x)} &= \pm \frac{1}{\sqrt{2}} \\ x &= \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} + 2\pi n, n \in \mathbf{Z} \end{align*} $

So the solution to the equation $\displaystyle \displaystyle \begin{align*} \sin{(5x)} + \sin{(x)} = 0 \end{align*} $ is

$\displaystyle \displaystyle \begin{align*} x = \left\{ 0, \frac{\pi}{4}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{4} \right\} + 2\pi n, n \in \mathbf{Z} \end{align*} $

Re: Trig equation: sin(5x)+sin(x)=0

Thank you very much for both of the replies but I know these methods.

The reason I set up the thread is to check if the method I gave a) is correct so far and b) can be used to find all the solutions.

Quote:

Originally Posted by

**earboth** **hmmmm, I never heard about that identity**

The identity I used was due to the fact that sin(x) is an odd function.

Re: Trig equation: sin(5x)+sin(x)=0

I realised that my method would miss some solutions. Instead it should be:

$\displaystyle \displaystyle \sin 5(x+(-1)^{k_1} k_1 \pi) = \sin -(x+(-1)^{k_2} k_2 \pi)$

And this will make things too confusing when the inequalities are produced. I think I'll give up on this now.

Re: Trig equation: sin(5x)+sin(x)=0

Actually there's a much simpler way to do it:

$\displaystyle \displaystyle \sin(a)=\sin(b) \implies a=b+2k_1 \pi$ or $\displaystyle a=\pi-b+2k_2 \pi$

where $\displaystyle k_1$ and $\displaystyle k_2$ are integers.

Subbing in $\displaystyle a=5x$ and $\displaystyle b=x$ leads to the solutions much faster than using the methods given by other people.

Re: Trig equation: sin(5x)+sin(x)=0

The simplest way I can think of to solve this is to use the general solution for trigonometric equations, which in this case is

If

$\displaystyle sinx &= b$

then

$\displaystyle x &= n\pi + \left(-1\right)^{n}sin^{-1}\left(b\right)$

This gives you all possible solutions for $\displaystyle x$ for integer $\displaystyle n$.

So, the problem looks like this:

$\displaystyle \\sin5x + sinx &= 0\\sin5x &= -sinx\\sin5x &= sin\left(-x\right)\\5x &= n\pi + \left(-1\right)^{n}\left(-x\right)$

If you now start plugging in values for $\displaystyle n$ from 0 onwards you should end up with the following valid answers (obviously restricted by the range you are given for $\displaystyle x$).:

$\displaystyle 0, \frac{\pi}{4}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{3\pi}{4}$

I hope that's helpful.

Re: Trig equation: sin(5x)+sin(x)=0

THere's the trig identity

$\displaystyle \sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}$,

so the equation can be rewritten as $\displaystyle 2sin(3x)\cos(2x)=0$

so either $\displaystyle \sin(3x)=0,$ or $\displaystyle \cos(2x)=0.$