How do you solve #sqrt(3) - 2Sin(6x) = 0# from #[0,2pi]#?

1 Answer
Alan P.
Jul 24, 2015

#sqrt(3)-2sin(6x)=0# with #x epsilon [0,2pi]#
#rarr##color(white)("XXXX")##x = pi/9 or pi/18#

Explanation:

Temporarily replace #6x# with #theta#

#sqrt(3) - 2sin(theta) = 0#

#rarr##color(white)("XXXX")##sin( theta) = sqrt(3)/2#

within the range #[0,2pi]# this is one of the standard angles with
#color(white)("XXXX")##theta = pi/3# or #theta= (2pi)/3#

Since #theta = 6x#
#color(white)("XXXX")##6x = pi/3# or #6x = (2pi)/3#

#rarr##color(white)("XXXX")##x = pi/18# or #x = pi/9#

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