How can I solve this?

#2sin2x - sqrt2 =0#

1 Answer
Hriman
Mar 13, 2018

#x= pi/8+pin#
#x=(3pi)/8+pin#

Explanation:

This is one of the multi angle problems:

#2sin2x-sqrt2=0#
#2sin2x=sqrt2#
#sin2x=sqrt2/2#

#2x= pi/4+2pin#
#2x=(3pi)/4+2pin#

#x= pi/8+pin#
#x=(3pi)/8+pin#

Where n is an element of all integers

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