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

Saturday, September 6, 2008

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

using the identity $\displaystyle{\cos{{\left({2}{x}\right)}}}={2}{{\cos}}^{{2}}{\left({x}\right)}-{1}$

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

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

Let cos(x)=y,

$\displaystyle{2}{{y}}^{{2}}-{y}-{1}={0}$

solving using the quadratic formula,

$\displaystyle{y}=\frac{{{1}\pm\sqrt{{{{\left(-{1}\right)}}^{{2}}-{4}\cdot{2}{\left(-{1}\right)}}}}}{{{2}\cdot{2}}}$

$\displaystyle{y}=\frac{{{1}\pm\sqrt{{{9}}}}}{{4}}=\frac{{{1}\pm{3}}}{{4}}={1},-\frac{{1}}{{2}}$

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

cos(x)=-1/2

General solutions are,

$\displaystyle{x}=\frac{{{2}\pi}}{{3}}+{2}\pi{n},{x}=\frac{{{4}\pi}}{{3}}+{2}\pi{n}$

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

$\displaystyle{x}=\frac{{{2}\pi}}{{3}},{x}=\frac{{{4}\pi}}{{3}}$

cos(x)=1

General solutions are,

$\displaystyle{x}={0}+{2}\pi{n}$

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

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

combine all the solutions ,

$\displaystyle{x}={0},{x}={2}\pi,{x}=\frac{{{2}\pi}}{{3}},{x}=\frac{{{4}\pi}}{{3}}$

Blog