Q (9)  Find the general solution of the following equation

   \small \sin x + \sin 3x + \sin 5x = 0

Answers (1)

We know that
                      \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
We use this identity in our problem
                      \sin 5x + \sin x = 2\sin\frac{5x+x}{2}\cos\frac{5x-x}{2} =2\sin3x\cos2x
Now our problem simplifeis to
                           2\sin3x\cos2x+ \sin3x = 0
take sin3x common
                           \sin3x(2\cos2x+ 1) = 0
So, either 
            sin3x = 0                                or                           \cos2x = -\frac{1}{2}                \left ( \cos2x = -\cos\frac{\pi}{3} = \cos\left ( \pi - \frac{\pi}{3} \right ) = \cos\frac{2\pi}{3} \right )
                 3x = n\pi                                                             2x = 2n\pi \pm \frac{2\pi}{3}                                                        
                    x = \frac{n\pi}{3}                                                               x = n\pi \pm \frac{\pi}{3}
Where n \ \epsilon \ Z
                          

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