4cos^2(x-[tex] \frac{ \pi }{6} [/tex]) - 3 = 0. Created by krokket. algebra-ru

или или n∈Z или k∈Z n∈Z или k∈Z n∈Z или k∈Z

tex]) - 3 = 0...

krokket @krokket

July 2022 1 21 Report

4cos^2(x-[tex] \frac{ \pi }{6} [/tex]) - 3 = 0

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chamomile03

Verified answer

$4cos^2(x- \frac{ \pi }{6} )-3=0 \\ \\ (2cos(x- \frac{ \pi }{6}) )^2-( \sqrt{3})^2=0 \\ \\ (2cos(x- \frac{ \pi }{6})- \sqrt{3} )(2cos(x- \frac{ \pi }{6})+ \sqrt{3} )=0$

$2cos(x- \frac{ \pi }{6})- \sqrt{3}=0$ или $2cos(x- \frac{ \pi }{6})+ \sqrt{3}=0$

$cos(x- \frac{ \pi }{6})= \frac{ \sqrt{3} }{2}$ или $cos(x- \frac{ \pi }{6})= -\frac{ \sqrt{3} }{2}$

$x- \frac{ \pi }{6}=бarccos\frac{ \sqrt{3} }{2} +2 \pi n,$ n∈Z или $x- \frac{ \pi }{6}=бarccos(-\frac{ \sqrt{3} }{2}) +2 \pi k,$ k∈Z

$x- \frac{ \pi }{6} =б \frac{ \pi }{6}+2 \pi n,$ n∈Z или $x- \frac{ \pi }{6} =б \frac{5 \pi }{6}+2 \pi k,$ k∈Z

$x= б\frac{ \pi }{6}+ \frac{ \pi }{6}+2 \pi n,$ n∈Z или $x= б\frac{5 \pi }{6}+ \frac{ \pi }{6}+2 \pi k,$ k∈Z

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krokket July 2022 | 0 Ответы

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