Question

tg3x+tg2x+tgx=0
X€(apartina)R

Answer 1

[tex]tgx+tg2x+tg3x=0 \Rightarrow tgx+tg2x+tg(x+2x) = 0 \Rightarrow \\ \\ \Rightarrow tgx+tg2x+ \dfrac{tgx+tg2x}{1-tgx\cdot tg2x} = 0 \Rightarrow \\ \\ \Rightarrow (tgx+tg2x)\cdot\Big(1+ \dfrac{1}{1-tgx\cdot tg2x}\Big)=0 \Rightarrow \\ \\ \Rightarrow (tgx+tg2x)\cdot\Big( \dfrac{1-tgx\cdot tg 2x+1}{1-tgx\cdot tg2x}\Big) = 0 \Big|\cdot(1-tgx\cdot tg2x) \Rightarrow \\ \\ (tgx+tg2x)\cdot(2-tgx\cdot tg2x)=0 \\ \\ \boxed{1} \quad tgx+tg2x=0 \\ \\ $sau$ \\ \\ \boxed{2} \quad tgx\cdot tg2x=2 [/tex]

$\boxed{1}\quad tgx+tg2x=0 \Rightarrow tgx+tg(x+x) = 0 \Rightarrow \\ tgx+ \dfrac{tgx+tgx}{1-tgx\cdot tgx}=0 \Rightarrow tgx+ \dfrac{2tgx}{1-tg^2x} = 0 \Rightarrow \\ \\ \Rightarrow tgx\cdot \Big(1+ \dfrac{2}{1-tg^2x}\Big) = 0 \Rightarrow tgx\cdot\Big( \dfrac{1-tg^2x+2}{1-tg^2x} \Big)=0 \Rightarrow \\ \\ \Rightarrow tgx\cdot(3-tg^2x)=0 \\ \\ \boxed{a_1} \quad tgx=0 \Rightarrow x=k\pi, \quad k\in \mathbb_{Z} \\ \\ sau$

$\\ \\ \boxed{a_2} \quad tg^2x=3 \Rightarrow tgx = \pm \sqrt{3} \Rightarrow x=k\pi \pm \dfrac{\pi}{3},\quad k\in \mathbb_{Z} \\ \\ \\ \boxed{2} \quad tgx\cdot tg2x=2 \Rightarrow \\ ..... (aici la fel, folosesti formula tg(a+b) =... pentru tg2x) \\ (si aici e mult de munca....)$