Помогите пожалуйста!!)) а) √3tg^2x-3tg x=0 б) 2cos^2x+cos x-1=0. Created by ssuhodolskaa34. algebra-ru

Помогите пожалуйста!!)) а) √3tg^2x-3tg x=0 б) 2cos^2x+cos x-1=0...

ssuhodolskaa34 @ssuhodolskaa34

July 2022 1 3 Report

Помогите пожалуйста!!))
а) √3tg^2x-3tg x=0
б) 2cos^2x+cos x-1=0

Answers & Comments

NNNLLL54

Verified answer

$1)\; \; \sqrt3tg^2x-3tgx=0\\\\tgx(\sqrt3tgx-3)=0\\\\a)\; \; tgx=0\; ,\; \; x=\pi n,\; n\in Z\\\\b)\; \; \sqrt3tgx-3=0\; \; ,\; \; tgx=\sqrt3\; \; ,\; \; x=\frac{\pi}{3}+\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=\pi n\; \; ,\; \; x=\frac{\pi}{3}+\pi k\; ,\; n,k\in Z\; .$

$2)\; \; 2cos^2x+cosx-1=0\\\\t=cosx\; \; ,\; \; -1\leq cosx\leq 1\\\\2t^2+t-1=0\; \; ,\; \; D=9\; ,\; t_1=-1\; ,\; \; t_2=\frac{1}{2}\\\\a)\; \; cosx=-1\; ,\; \; x=\pi +2\pi k\; ,\; k\in Z\\\\b)\; \; cosx=\frac{1}{2}\; ,\; \; x=\pm \frac{\pi}{6}+\pi n\; ,\; n\in Z\\\\Otvet:\; \; x=\pi +2\pi k\; \; ,\; \; x=\pm \frac{\pi}{6}+2\pi n\; ,\; n,k\in Z\; .$

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