How do you solve #2cosx + sin2x = 0# in the interval [0, 2pi]?

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Answer 1 · 2016-07-05T13:13:59

#x= pi/2 or (3pi)/2#

#2cosx+sin2x=0#

#=>2cosx+2sinxcosx=0#

#=>2cosx(1+sinx)=0#

#=>cosx=0 and 1+sinx=0#

when
#cosx=0=cos(pi/2)=cos(2pi-pi/2)#

#:.x= pi/2 or x=(3pi)/2#

Again when

#1+sinx=0#

#=>sinx=-1=-sin(pi/2)=sin(pi+pi/2)#

#x=(3pi)/2#