Question

The general solution of $\sin \mathrm{\theta }+\sin 2\mathrm{\theta }+\sin 3\mathrm{\theta }+\sin 4\mathrm{\theta }=0$?

Answer 1

Given,

$\sin \mathrm{\theta }+\sin 2\mathrm{\theta }+\sin 3\mathrm{\theta }+\sin 4\mathrm{\theta }=0$

By rearranging the expression, we get

$(\mathrm{Sin}\mathrm{\theta }+\mathrm{Sin}3\mathrm{\theta })+(\sin 2\mathrm{\theta }+\mathrm{Sin}4\mathrm{\theta })=0$

$\begin{array}{rcl}(\sin \mathrm{\theta }+\sin 3\mathrm{\theta })+(\sin 2\mathrm{\theta }+\sin 4\mathrm{\theta })& =& 0\\ 2.\sin 2\mathrm{\theta }.\cos \mathrm{\theta }+2.\sin 3\mathrm{\theta }.\cos \mathrm{\theta }& =& 0\\ 2.\cos \mathrm{\theta }(\sin 2\mathrm{\theta }+\sin 3\mathrm{\theta })& =& 0\\ 4.\cos \mathrm{\theta }.\sin \frac{5\mathrm{\theta }}{2}.\cos \frac{\mathrm{\theta }}{2}& =& 0\end{array}$

$\begin{array}{rcl}\cos \mathrm{\theta }& =& 0=\cos \frac{\mathrm{\pi }}{2},\\ \mathrm{\theta }& =& (\mathrm{n}+\frac{1}{2})\mathrm{\pi }\end{array}$ $$\begin{gathered} \cos \frac{\mathrm{\theta }}{2}=0=\cos \frac{\mathrm{\pi }}{2} \\ \frac{\mathrm{\theta }}{2}=\left(\mathrm{n}+\frac{1}{2}\right)\mathrm{\pi } \\ \mathrm{\theta }=(2\mathrm{n}+1)\mathrm{\pi } \end{gathered}$$ $\begin{array}{rcl}\sin \frac{5\mathrm{\theta }}{2}& =& 0=\sin 0\\ \frac{5\mathrm{\theta }}{2}& =& \mathrm{n\pi }\\ \mathrm{\theta }& =& \frac{2\mathrm{n\pi }}{5}\end{array}$

Hence, the general solution of $\sin \mathrm{\theta }+\sin 2\mathrm{\theta }+\sin 3\mathrm{\theta }+\sin 4\mathrm{\theta }=0$ as shown above.