Вычислить: 2arcsin√3/2+3arcsin(-1/2) arccos(-1/2)-arcsin√3 Решить: cos(4-2x)=-1/2 √2cos(2х+п/4)+1=0. Created by lyricslizergin. algebra-ru

4)+1=0...

lyricslizergin @lyricslizergin

August 2022 1 8 Report

Вычислить:

2arcsin√3/2+3arcsin(-1/2)

arccos(-1/2)-arcsin√3

Решить:

cos(4-2x)=-1/2

√2cos(2х+п/4)+1=0

Answers & Comments

NNNLLL54

Verified answer

$1)\; \; 2arcsin\frac{\sqrt3}{2}+3arcsin(-\frac{1}{2})=2\cdot \frac{\pi}{3}+3\cdot (-\frac{\pi}{6})=-\frac{\pi}{6}\\\\2)\; \; arccos(-\frac{1}{2})-arcsin\sqrt3=(\pi -\frac{\pi}{3})-arcsin\sqrt3=\frac{2\pi}{3}-arcsin\sqrt3\\\\3)\; \; cos(4-2x)=-\frac{1}{2}\\\\4-2x=\pm arccos(-\frac{1}{2})+2\pi n=\pm (\pi -\frac{\pi}{3})+2\pi n=\pm \frac{2\pi}{3}+2\pi n\\\\-2x=\pm \frac{2\pi}{3}-4+2\pi n\; ,\; \; n\in Z\\\\x=\mp\frac{\pi}{3}+2+\pi n\; ,\; \; n\in Z\\\\4)\; \; \sqrt2cos(2x+\frac{\pi}{4})+1=0$

$cos(2x+\frac{\pi}{4})=-\frac{\sqrt2}{2}\\\\2x+\frac{\pi}{4}=\pm arccos(-\frac{\sqrt2}{2})+2\pi n=\pm \frac{3\pi }{4}+2\pi n,\; n\in Z\\\\2x=\pm \frac{3\pi}{4}-\frac{\pi}{4}+2\pi n\; ,\; n\in Z\\\\x=\pm \frac{3\pi }{8}-\frac{\pi}{8}+\pi n\; ,\; n\in Z$

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