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

Sunday, 28 May 2017

$\displaystyle{\tan{{\left({2}{x}\right)}}}-{2}{\cos{{\left({x}\right)}}}={0}$ Find the exact solutions of the equation in the interval [0, 2pi).

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

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

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

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

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

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

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

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

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

solving each part separately,

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

General solutions are,

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

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

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

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

Let sin(x)=y

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

solve using the quadratic formula,

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

$\displaystyle{y}=-{1},\frac{{1}}{{2}}$

substitute back y=sin(x)

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

For sin(x)=-1

General solutions are,

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

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

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

For sin(x)=1/2

General solutions are,

$\displaystyle{x}=\frac{\pi}{{6}}+{2}\pi{n},{x}=\frac{{{5}\pi}}{{6}}+{2}\pi{n}$

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

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

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

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

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

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

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

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

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

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

solving each part separately,

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

General solutions are,

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

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

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

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

Let sin(x)=y

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

solve using the quadratic formula,

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

$\displaystyle{y}=-{1},\frac{{1}}{{2}}$

substitute back y=sin(x)

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

For sin(x)=-1

General solutions are,

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

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

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

For sin(x)=1/2

General solutions are,

$\displaystyle{x}=\frac{\pi}{{6}}+{2}\pi{n},{x}=\frac{{{5}\pi}}{{6}}+{2}\pi{n}$

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

$\displaystyle{x}=\frac{\pi}{{6}},\frac{{{5}\pi}}{{6}}$

Combine all the solutions,

$\displaystyle{x}=\frac{\pi}{{2}},\frac{{{3}\pi}}{{2}},\frac{\pi}{{6}},\frac{{{5}\pi}}{{6}}$

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