Question

If \[\sin x + \sin 3x + \sin 5x = 0\] , then the solution is?

Answer 1

Hint:
We are given \[sinx + sin3x + sin5x = 0\] and we are trying to find the value of x. We use the formula of \[\sin a + \sin b = 2\sin \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}\] on \[\sin 5x + \sin x\] we get answer in terms of \[\sin 3x\]. Next, we take \[\sin 3x\]common and then equating it with zero we find our solution.

Complete step by step solution:
 \[\sin x + \sin 3x + \sin 5x = 0\]
Taking two terms together, we get,
 \[ \Rightarrow \left( {\sin 5x + \sin x} \right) + \sin 3x = 0\]
Again, we have, \[\sin a + \sin b = 2\sin \dfrac{{a + b}}{2}\cos \dfrac{{a - b}}{2}\] , using, this we get,
 \[ \Rightarrow 2sin3xcos2x + sin3x = 0\]
Taking, \[\sin 3x\] common, we get,
 \[ \Rightarrow sin3x\left( {2cos2x + 1} \right) = 0\]
Now,
 \[\sin 3x = 0\;or\;2\cos 2x + 1 = 0\]
 \[ \Rightarrow \sin 3x = 0\;or,\; \cos 2x = - \dfrac{1}{2}\]
Now, \[\;\sin 3x = 0 \Rightarrow 3x = n\pi \Rightarrow x = \dfrac{{n\pi }}{3},n \in Z\] as, \[\sin n\pi = 0\]
And, \[\;\cos 2x = - \dfrac{1}{2}\]
 \[\cos 2x = \cos \dfrac{{2\pi }}{3}\] as, \[\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}\]
 \[ \Rightarrow 2x = 2m\pi \pm \dfrac{{2\pi }}{3},m \in Z\]
 \[ \Rightarrow x = m\pi \pm \dfrac{\pi }{3},m \in Z\]

Hence, the general solution of the given equation is:
\[x = \dfrac{{n\pi }}{3}\]​ or, \[x = m\pi \pm \dfrac{\pi }{3}\] , where \[m,n \in Z\]

Note:
To find, \[\sin n\pi = 0\] , we get,
Recall that Euler's formula is \[{e^{ix}} = cosx + isinx\] , When \[x = \pi \] ,
we have \[{e^{i\pi }} = cos\pi + isin\pi = - 1\] \[ \Rightarrow {e^{i\pi }} + 1 = 0\]
This implies that \[\sin n\pi = 0\] for all \[n \in \mathbb{Z}\].