Question

Помогите решить уравнение!

Answer 1

Ответ:

$-arctg4+\pi k, \ k \in \mathbb{Z} \ ; \quad \dfrac{\pi}{4}+\pi k, \ k \in \mathbb{Z} \ ;$

Пошаговое объяснение:

$D(x): \ \cos^{2}x \neq 0 \Rightarrow \cos x \neq 0 \Rightarrow x \neq\dfrac{\pi}{2}+2\pi k, \ k \in \mathbb{Z};$

$\dfrac{1}{\cos^{2}x}+3tg x-5=0;$

$tg^{2}x+1+3tgx-5=0;$

$tg^{2}x+3tgx-4=0;$

$tg^{2}x-tgx+4tgx-4=0;$

$tgx(tgx-1)+4(tgx-1)=0;$

$(tgx+4)(tgx-1)=0;$

$tgx+4=0 \quad \vee \quad tgx-1=0;$

$tgx=-4 \quad \vee \quad tgx=1;$

$x=arctg(-4)+\pi k, \ k \in \mathbb{Z} \quad \vee \quad x=arctg1+\pi k, \ k \in \mathbb{Z};$

$x=-arctg4+\pi k, \ k \in \mathbb{Z} \quad \vee \quad x=\dfrac{\pi}{4}+\pi k, \ k \in \mathbb{Z} \ ;$

Оба корня удовлетворяют ОДЗ.

Answer 2

$\displaystyle\frac{1}{cos^2x}+3tgx-5=0\\\\\\ODZ:\ cos^2x\ne0\\\\cosx\ne0\\\\\boxed{x\ne\frac{\pi}{2}+2\pi n,\ n\in Z}\\\\\\----------\\\\\\\boxed{cos^2x=\frac{1}{tg^2x+1}\Rightarrow\dfrac{1}{cos^2x}=tg^2x+1}\\\\\\tg^2x+1+3tgx-5=0\\\\tg^2x+3tgx-4=0\\\\\boxed{tgx=t}\\\\t^2+3t-4=0\\\\D=3^2-4\cdot(-4)=9+16=25\\\\t_1=\dfrac{-3-5}{2}=\dfrac{-8}{2}=-4\\\\\\t_2=\dfrac{-3+5}{2}=\dfrac{2}{2}=1$

$1)\ tgx=-4\\\\x=arctg(-4)\\\\\boxed{x=-arctg4+\pi n,\ n\in Z};\ x\ne\dfrac{\pi}{2}+\pi n,\ n\in Z\\\\\\2)\ tgx=1\\\\x=arctg1\\\\\boxed{x=\dfrac{\pi}{4}+\pi n,\ n\in Z};\ x\ne\dfrac{\pi}{2}+\pi n,\ n\in Z$

Ответ:

$x=-arctg4+\pi n,\ n\in Z\\\\x=\dfrac{\pi}{4}+\pi n,\ n\in Z$