How do you solve #2sinx = sqrt2#?

1 Answer
Binayaka C.
Oct 15, 2016

General solution : # x= (2n*pi+pi/4) and x= (2n*pi+(3pi)/4)#

Explanation:

#2sinx = sqrt 2 or sinx =sqrt2/2#; In interval #[0,2pi] sin (pi/4)=sqrt2/2 and sin(pi-pi/4)=sqrt2/2 :. x= pi/4 and x= (3pi)/4#
General solution : # x= (2n*pi+pi/4) and x= (2n*pi+(3pi)/4)# Where #n# is an integer.[Ans]

Related questions
Impact of this question
8651 views around the world
You can reuse this answer
Creative Commons License