Question

$3\tan ^3\left(A\right)-\tan \left(A\right)=0,\:A\in \left[0,\:360\right]$

solve it !!!!!!!!!!!!!!!!!!!!!

Answer 1

$\mathrm{Let:\:}\tan \left(A\right)=u$

$3u^3-u=0\quad :\quad u=0,\:u=-\frac{\sqrt{3}}{3},\:u=\frac{\sqrt{3}}{3}$

$\tan \left(A\right)=0,\:\tan \left(A\right)=-\frac{\sqrt{3}}{3},\:\tan \left(A\right)=\frac{\sqrt{3}}{3}$

$\tan \left(A\right)=0\quad :\quad A=\pi n$$\tan \left(A\right)=-\frac{\sqrt{3}}{3}\quad :\quad A=\frac{5\pi }{6}+\pi n$

$A=\pi n,\:A=\frac{5\pi }{6}+\pi n,\:A=\frac{\pi }{6}+\pi n$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\bf{3\tan ^3\left(A\right)-\tan \left(A\right)=0,\:A\in \:\left[0,\:360\right]}$

♣ ᴀɴꜱᴡᴇʀ :

$\bf{Let:\:}\tan \left(A\right)=u}$

$\bf{3u^3-u=0}$

$\bf{3u^3-u=0\quad :\quad u=0,\:u=-\dfrac{\sqrt{3}}{3},\:u=\dfrac{\sqrt{3}}{3}}$

$\bf{Substitute\:back\:u=\tan \left(A\right)}$

$\bf{\tan \left(A\right)=0,\:\tan \left(A\right)=-\dfrac{\sqrt{3}}{3},\:\tan \left(A\right)=\dfrac{\sqrt{3}}{3}}$

$\rule{250}{1}$

$\bf{\tan \left(A\right)=0,\:\tan \left(A\right)=-\dfrac{\sqrt{3}}{3},\:\tan \left(A\right)=\dfrac{\sqrt{3}}{3}}$

$\bf{tan \left(A\right)=-\dfrac{\sqrt{3}}{3},\:0\le \:A\le \:360^{\circ \:}\quad :\quad A=150^{\circ \:},\:A=330^{\circ \:}}$

$\bf{\tan \left(A\right)=\frac{\sqrt{3}}{3},\:0\le \:A\le \:360^{\circ \:}\quad :\quad A=30^{\circ \:},\:A=210^{\circ \:}}$

$\rule{250}{1}$

$\bf{Combine\:all\:the\:solutions}$

$\boxed{\bf{A=0,\:A=180^{\circ \:},\:A=360^{\circ \:},\:A=150^{\circ \:},\:A=330^{\circ \:},\:A=30^{\circ \:},\:A=210^{\circ \:}}}$