Solution assignment 07 Addition and subtraction formulas

Solve for $0\leq{x}\leq{2\pi}$ :

$\sin(5x)+\sin(x)-\sin(3x)=0$

The first two terms can be written with Simpson/Mollweide as:

$\sin(5x)+\sin(x)=2\sin(3x)\cos(2x)$

and thus the equation becomes:

$2\sin(3x)\cos(2x)-\sin(3x)=0$

$\sin(3x)[2\cos(2x)-1]=0$

One factor yields:

$\sin(3x)=0$

meaning:

$x=\displaystyle\frac{1}{3}\pi, \displaystyle\frac{2}{3}\pi, ..., 2\pi$

The other factor yields:

$2\cos(2x)-1=0$

$\cos(x)=\displaystyle\frac{1}{2}$

thus:

$x=\displaystyle\frac{1}{12}\pi, \displaystyle\frac{13}{12}\pi$