Verified answer

Ответ:

[tex]\left [ \begin{array} {l} x=-\dfrac{\pi }{20} +\cfrac{\pi n }{5} \\\\ x=\cfrac{ \mathrm{arctg}(-3)+\pi n }{5} ~ ; ~ n \in \mathbb {Z} \end{array} \right.[/tex]

Объяснение:

[tex]\displaystyle \mathrm {tg }5x+3 \mathrm{ctg}5x+4=0 \\\\\\ \mathrm {tg }5x+\frac{3}{ \mathrm{tg}5x} +4 =0 ~~ \big |\cdot \mathrm{tg}5x \\\\\\ \mathrm {tg^2 }5x+4 \mathrm{tg}5x +3 =0[/tex]

Сделаем замену

[tex]\mathrm{tg}5x = t ~~ ; ~~ \mathrm{tg^2}5x =t^2[/tex]

[tex]t^2+4t+3 =0 \\\\ \displaystyle \left \{ \begin{array}{l} t_1+t_2=-4 \\\\t_1t_2=3\end{array} \right. \Leftrightarrow t_1= -1 ~~ ;~~ t_2=-3[/tex]

[tex]\hspace{-1em} 1)~ \mathrm{tg}5x = -1 \\\\ 5x = \mathrm{arctg}(-1) +\pi n\\\\ 5x =- \cfrac{\pi }{4} + \pi n \\\\ x=-\dfrac{\pi }{20} +\cfrac{\pi n }{5} ~~ ;~~ n \in \mathbb {Z}[/tex]


[tex]\hspace{-1em} 2)~ \mathrm{tg}5x = -3 \\\\ 5x = \mathrm{arctg}(-3)+\pi n \\\\ x=\cfrac{ \mathrm{arctg}(-3)+\pi n }{5} ~ ; ~ n \in \mathbb {Z}[/tex]