Sin3x - sinx = 0 (for x greater than 0 but less than 360)

1. Feb 2, 2008

Rush147

Hi folks, can anybody help me. I would like to solve the following equation: ""sin3x - sinx = 0 (for x greater than 0 but less than 360) and come up with angles for x or 2x. We are doing Trigonometrical identities which i have only just come across. Got as far as "sin(2x + x) - sinx = 0" as we think we need to get a 2x or x and sin(2x + x) is also an identity that can be changed, but not sure if this is the right direction.

Many thanks folks

2. Feb 2, 2008

neutrino

Can you write the x (in the second term) in terms of 2x and x like you did for 3x?

3. Feb 2, 2008

Rush147

I believe so as this gives "sin(a + b)" (or in my case sin(2x + x)) and this is equal (or can then be substituted with the identity) "sinacosb + cosasinb", BUT i really don't know if i'm heading the correct way as i've only just this week come across trigonometrical identities so it could be totally different. I know the last equation i did gave me a double angle (2x) which gave 2 angles within 360 degrees. Any help you can give would be most appreciated.

4. Feb 2, 2008

neutrino

I meant this: sin(2x+x) - sin(2x-x) = 0.

5. Feb 2, 2008

scottie_000

Well if you want to try it this way let a=2x and b=x, then use that same formula again on the terms that have 2x's in them. You only want sines from then on....
But it's a lot of algebra

6. Feb 2, 2008

Rush147

I don't think so as sin(2x+x) could be a backwards step from sin3x but with the "- sinx" there is only one x there so not sure where the 2x and x would come from.

7. Feb 2, 2008

VietDao29

There are actually 3 ways to solve the above problem.

1. The first way, the most straightforward, and require the most calculation is to trace 3x down to x, by using Triple-Angle Formulae: sin(3x) = 3 sin(x) - 4sin3(x) (You can arrive to this formula by using the Sum-Angle Identity twice). So, your equation becomes a cubic equation in sin(x), which is pretty easy to solve. :)

2. The second way, the easiest way, is to use the properties of sin function:

$$\sin \alpha = \sin \beta$$

$$\Leftrightarrow \left[ \begin{array}{lcr} \alpha & = & \beta + k 360 ^ o \\ \alpha & = & 180 ^ o - \beta + k' 360 ^ o \end{array} \right.$$, k, and k' are both integers.

One can isolate sin(x) to the other side of the equation: sin(3x) = sin(x), and use method mentioned above. Then choose, k, and k' wisely so that your solution is on the interval [0, 360]

3. The 3rd way, the final one, is to use the Sum-To-Product Identities:
(You can arrive to this Itentity by using neutrino's hint)
$$\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)$$

Hopefully, you can go from here. :)

You can pick up one of the 3 ways mentioned above, or try all 3, and compare the result. :)

Last edited: Feb 2, 2008
8. Feb 2, 2008

Rush147

I'm not sure if thats at all the correct way but a friend of mine started by using sin(a + b) as this was an identity that could be replaced. These trig identities are all new to me and not the best at algebra either

9. Feb 2, 2008

Rush147

Thanks for your help. Its all very confusing as i'm quite new to this level of maths. I will have a look through.