How do I solve #2sin^2x-1=0#?

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Answer 1 · 2018-06-07T21:56:57

#pi/4, (3pi)/4, (5pi)/4, (7pi)/4# on the range #(0-2pi)#

or to be more exact:

#pi/4 + k*pi/2# where #k# is all integers.

#2sin^2x-1=0#

#2sin^2x=1#

#sin^2x=1/2#

#sinx=+-sqrt(1/2)#

#sinx=+-sqrt(2)/2#

Using the unit circle we see:

#sinx=sqrt(2)/2# at #pi/4 and (3pi)/4#

#sinx=-sqrt(2)/2# at #(5pi)/4# and #(7pi)/4#

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