neudateaLp

Answered

2022-11-25

Find all solutions to $\mathrm{tan}(x-\pi )+2\mathrm{sin}(x-\pi )=0$ on $[0,2\pi )$

Answer & Explanation

ysik92ASw

Expert

2022-11-26Added 13 answers

Given

$\mathrm{tan}(x-\pi )+2\mathrm{sin}(x-\pi )=0$

We know that

$\frac{\mathrm{sin}(x-\pi )}{\mathrm{cos}(x-\pi )}+2\mathrm{sin}(x-\pi )=0\phantom{\rule{0ex}{0ex}}\mathrm{sin}(x-\pi )[\frac{1}{\mathrm{cos}(x-\pi )}+2]=0\phantom{\rule{0ex}{0ex}}\text{If}\mathrm{sin}(x-\pi )=0\phantom{\rule{0ex}{0ex}}x-\pi ={\mathrm{sin}}^{-1}(0)=0\phantom{\rule{0ex}{0ex}}x=\pi $

If $\frac{1}{\mathrm{cos}(x-\pi )}+2=0\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{cos}(x-\pi )}=-2\phantom{\rule{0ex}{0ex}}\mathrm{cos}(x-\pi )=\frac{1}{-2}\phantom{\rule{0ex}{0ex}}x-\pi ={\mathrm{cos}}^{-1}(\frac{1}{-2})\phantom{\rule{0ex}{0ex}}x-\pi ={60}^{\circ}\phantom{\rule{0ex}{0ex}}x={240}^{\circ}\phantom{\rule{0ex}{0ex}}x=\frac{4\pi}{3}$

$\mathrm{tan}(x-\pi )+2\mathrm{sin}(x-\pi )=0$

We know that

$\frac{\mathrm{sin}(x-\pi )}{\mathrm{cos}(x-\pi )}+2\mathrm{sin}(x-\pi )=0\phantom{\rule{0ex}{0ex}}\mathrm{sin}(x-\pi )[\frac{1}{\mathrm{cos}(x-\pi )}+2]=0\phantom{\rule{0ex}{0ex}}\text{If}\mathrm{sin}(x-\pi )=0\phantom{\rule{0ex}{0ex}}x-\pi ={\mathrm{sin}}^{-1}(0)=0\phantom{\rule{0ex}{0ex}}x=\pi $

If $\frac{1}{\mathrm{cos}(x-\pi )}+2=0\phantom{\rule{0ex}{0ex}}\frac{1}{\mathrm{cos}(x-\pi )}=-2\phantom{\rule{0ex}{0ex}}\mathrm{cos}(x-\pi )=\frac{1}{-2}\phantom{\rule{0ex}{0ex}}x-\pi ={\mathrm{cos}}^{-1}(\frac{1}{-2})\phantom{\rule{0ex}{0ex}}x-\pi ={60}^{\circ}\phantom{\rule{0ex}{0ex}}x={240}^{\circ}\phantom{\rule{0ex}{0ex}}x=\frac{4\pi}{3}$

Most Popular Questions