`sin(2x) - sin(x) = 0` Find the exact solutions of the equation in the interval [0, 2pi).

Wednesday, February 25, 2009

$\displaystyle{\sin{{\left({2}{x}\right)}}}-{\sin{{\left({x}\right)}}}={0},{0}\le{x}\le{2}\pi$

using the identity $\displaystyle{\sin{{\left({2}{x}\right)}}}={2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}$

$\displaystyle{\sin{{\left({2}{x}\right)}}}-{\sin{{\left({x}\right)}}}={0}$

$\displaystyle={2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}-{\sin{{\left({x}\right)}}}={0}$

$\displaystyle={\left({\sin{{\left({x}\right)}}}\right)}{\left({2}{\cos{{\left({x}\right)}}}-{1}\right)}={0}$

solving each part separately,

$\displaystyle{\sin{{\left({x}\right)}}}={0}$ or $\displaystyle{2}{\cos{{\left({x}\right)}}}-{1}={0}$

General solutions for sin(x)=0 are,

$\displaystyle{x}={0}+{2}{\left(\pi\right)}{n},{x}=\pi+{2}{\left(\pi\right)}{n}$

solutions for the range $\displaystyle{0}\le{x}\le{2}\pi$ are,

$\displaystyle{x}={0},{x}=\pi,{x}={2}\pi$

Now solving 2cos(x)-1=0,

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

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

General solutions for cos(x)=1/2 are,

$\displaystyle{x}=\frac{{\pi}}{{3}}+{2}\pi{n},{x}=\frac{{{5}\pi}}{{3}}+{2}\pi{n}$

solutions for the range $\displaystyle{0}\le{x}\le{2}\pi$ are,

$\displaystyle{x}=\frac{\pi}{{3}},{x}=\frac{{{5}\pi}}{{3}}$

Hence all the solutions are,

$\displaystyle{x}={0},\pi,{2}\pi,\frac{\pi}{{3}},\frac{{{5}\pi}}{{3}}$

Blog