Question 17.2.9: Find the general solution of the equation sin 2x+sin 4x+sin ......
Find the general solution of the equation \sin 2 x+\sin 4 x+\sin 6 x=0 .
Step-by-Step
The given equation may be written as (\sin 6 x+\sin 2 x)+\sin 4 x=0 \\ \\\begin{aligned}&\Rightarrow 2 \sin \frac{(6 x+2 x)}{2} \cos \frac{(6 x-2 x)}{2}+\sin 4 x =0 \\ \\& {\left[\because \sin C+\sin D=2 \sin \frac{(C+D)}{2} \cos \frac{(C-D)}{2}\right] }\end{aligned}\\ \\
\begin{array}{l}\Rightarrow \quad 2 \sin 4 x \cos 2 x+\sin 4 x=0 \\ \\\Rightarrow \quad \sin 4 x(2 \cos 2 x+1)=0 \\ \\\Rightarrow \quad \sin 4 x=0 \text { or } 2 \cos 2 x+1=0 \\ \\\Rightarrow \quad \sin 4 x=0 \text { or } \cos 2 x=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} \\ \\\Rightarrow \quad \sin 4 x=0 \text { or } \cos 2 x=\cos \frac{2 \pi}{3} \\ \\\Rightarrow \quad 4 x=n \pi \text { or } 2 x=\left(2 m \pi \pm \frac{2 \pi}{3}\right), \text { where } m, n \in I\\ \\\Rightarrow \quad x=\frac{n \pi}{4} \text { or } x=\left(m \pi \pm \frac{\pi}{3}\right), \text { where } m, n \in I .\end{array}
Hence, the general solution is given by x=\frac{n \pi}{4} or x=\left(m \pi \pm \frac{\pi}{3}\right) , where m, n \in I .
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