Find the general solution of sin (3x) + sin (x) = 0. The textbook says the answer is (pi*n)/2 but I can’t seem to get it. Please help?
do you know the formula
$\displaystyle sinA+sinB=2{sin(\frac{A+B}{2})cos(\frac{A-B}{2})}$
so
$\displaystyle sin3x+sinx=2sin(\frac{4x}{2})cos(\frac{2x}{2})=2si n(2x)cos(x)=0$
so sin2x=0 or cosx=0
sin2x=0
$\displaystyle sin2x=0.....2x=n\pi....x=\frac{n\pi}{2}$
$\displaystyle cosx=0.....x=\frac{n\pi}{2}$
A simple approach:
$\displaystyle \sin (3x) + \sin (x) = 0 \Rightarrow \sin (3x) = - \sin (x) = \sin (-x)$.
Case 1: $\displaystyle 3x = -x + 2 n \pi$ where n is an integer.
Case 2: $\displaystyle 3x = (\pi - (-x)) + 2 n \pi = x + (2n + 1) \pi$. The solutions here are a subset of those in Case 1.