Question #5869c

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Answer 1 · 2018-01-30T02:49:10

#sinx +(sin^2x)/cosx =0#

#=>sinx +sinxtanx =0#

#=>sinx(1 +tanx) =0#

So #sinx=0#

#x=npi " where "n in ZZ#

Again when #tanx+1=0#

#=>tanx=tan(-pi/4)#

#x=kpi -pi/4"where "k inZZ#