tg3x+tg2x−3tgx−3=0\tg^3 x + \tg^2 x - 3\tg x - 3 = 0tg3x+tg2x−3tgx−3=0 Показать решение Решение: (∗)cosx≠0⇔x≠π2+πk,k∈Z;\left(*\right) \cos x \neq 0 \Leftrightarrow x \neq \frac{\pi}{2} + \pi k, k \in Z;(∗)cosx=0⇔x=2π+πk,k∈Z; Сделаем замену t=tgxt = \tg xt=tgx: t3+t2−3t−3=0t^3 + t^2 - 3t - 3 = 0t3+t2−3t−3=0 (t3+t2)−(3t+3)=0(t^3 + t^2) - (3t + 3) = 0(t3+t2)−(3t+3)=0 t2(t+1)−3(t+1)=0t^2(t + 1) - 3(t + 1) = 0t2(t+1)−3(t+1)=0 (t+1)(t2−3)=0(t + 1)(t^2 - 3) = 0(t+1)(t2−3)=0 t=−1 или t2=3t = -1 \text{ или } t^2 = 3t=−1 или t2=3 tgx=−1⇒x=−π4+πk,k∈Z\tg x = -1 \Rightarrow x = -\frac{\pi}{4} + \pi k, \quad k \in \mathbb{Z}tgx=−1⇒x=−4π+πk,k∈Z tgx=±3⇒x=±π3+πk,k∈Z\tg x = \pm \sqrt{3} \Rightarrow x = \pm \frac{\pi}{3} + \pi k, \quad k \in \mathbb{Z}tgx=±3⇒x=±3π+πk,k∈Z