How do you solve #2sin^2x=1#?

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Luca F. Share
Jun 10, 2015

Answer:

#S={x|x=pi/4+k/2pi " where " kinRR}#.

Explanation:

#2sin^2x=1#
#sin^2x=1/2#
#sinx=+-sqrt(1/2)#
#sinx=+-sqrt(2)/2#
This is a common value that we get with #x=pi/4,3/4pi,5/4pi,7/4pi,...#
So, the solutions is:
#S={x|x=pi/4+k/2pi " where " kinRR}#.

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