`tan(x + pi) + 2sin(x + pi) = 0` Find all solutions of the equation in the interval [0, 2pi).

Friday, October 23, 2015

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

using $\displaystyle{\tan{{\left({x}+\pi\right)}}}={\tan{{\left({x}\right)}}},{\sin{{\left({x}+\pi\right)}}}=-{\sin{{\left({x}\right)}}}$

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

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

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

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

solving each part separately,

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

For sin(x)=0,

General solutions are,

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

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

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

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

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

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

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

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

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

combine all the solutions,

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

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