Question #c20c8

1 Answer
Isini M.
Jun 6, 2017

#cos^2x=1+sinx#
#sin^2x+cos^2x=1#
Therefore, #cos^2x=1-sin^2x#
So,
#1-sin^2x=1+sinx#
#sin^2x+sinx=0#
#sinx(sinx+1)=0#
Use null factor law (keeping in mind interval (0, 2π))
#sinx=0# and #sinx+1=0#

#sinx=0#
#x=0,2pi#

#sinx+1=0#
#sinx=-1#
#x=(3pi)/2#
Therefore, #x=0,2pi,(3pi)/2#

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