Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 11: a.Tanx=√3/3 b.cot3x=√3 Giải các phương trình sau

Toán Lớp 11: a.Tanx=√3/3
b.cot3x=√3
Giải các phương trình sau

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
     $a)đkxđ: x\neq \dfrac{\pi}{2}+k\pi \\x=\dfrac{\pi}{6}+k\pi  (k\in Z)$
    $b) đkxđ: x\neq \dfrac{k\pi}{3}\\3x= \dfrac{\pi}{6}+k\pi \Leftrightarrow x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}$

  2. ~rai~
    \(a)\tan x=\dfrac{\sqrt{3}}{3}\quad(1)\\ĐKXĐ:\cos x\ne 0\\\Leftrightarrow x\ne \dfrac{\pi}{2}+k\pi.(k\in\mathbb{Z})\\(1)\Leftrightarrow \tan x=\tan\dfrac{\pi}{6}\\\Leftrightarrow x=\dfrac{\pi}{6}+k\pi.(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{6}+k\pi\Big|k\in\mathbb{Z}\right\}.\\b)\cot3x=\sqrt{3}\quad(1)\\ĐKXĐ:\sin3x\ne 0\\\Leftrightarrow 3x\ne k\pi\\\Leftrightarrow x\ne k\dfrac{\pi}{3}.(k\in\mathbb{Z})\\(1)\Leftrightarrow \cot3x=\cot\dfrac{\pi}{6}\\\Leftrightarrow 3x=\dfrac{\pi}{6}+k\pi\\\Leftrightarrow x=\dfrac{\pi}{18}+k\dfrac{\pi}{3}.(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{18}+k\dfrac{\pi}{3}\Big|k\in\mathbb{Z}\right\}.\)

Leave a reply

222-9+11+12:2*14+14 = ? ( )

About Melanie