Question

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

Answer 1

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 $n \in \mathbb{Z}$.