How do you find all the solutions between 0 and 2π for #4sin^2x-3 = 0#?

1 Answer
Alan P.
Aug 17, 2017

#x in {pi/3, (2pi)/3, (5pi)/3, (7pi)/3}#

Explanation:

If #4sin^2x-3=0#
then
#color(white)("XXX")sin^2x=3/4#
and
#color(white)("XXX")sin x=+-sqrt(3)/2#

This is one of the standard reference angles #= pi/3#

Between 0 and #2pi#, the possibilities for this reference angle are
#color(white)("XXX")pi/3,(2pi)/3, (5pi)/3, (7pi)/3#

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