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

[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

 

[neutrino]

"sin(2x + x) - sinx = 0

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

 

[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.

 

[neutrino]

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

 

[scottie_000]

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"

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

 

[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.

 

[VietDao29]

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.

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. :)

 

[Rush147]

I'm not sure if that's 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

 

[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.