Given
#sin theta_1 + sin theta_2 + sin theta_3=3#
#=>1-sin theta_1 +1- sin theta_2 +1- sin theta_3=0#
#=>1-2sin (theta_1/2)cos(theta_1/2)+1- 2sin (theta_2/2)cos(theta_2/2) +1- 2sin (theta_3/2)cos(theta_3/2)=0#
#=>sin^2 (theta_1/2)+cos^2(theta_1/2)-2sin (theta_1/2)cos(theta_1/2)+sin^2 (theta_2/2)+cos^2(theta_2/2)- 2sin (theta_2/2)cos(theta_2/2) +sin^2 (theta_3/2)+cos^2(theta_3/2)- 2sin (theta_3/2)cos(theta_3/2)=0#
#=>(sin (theta_1/2)-cos(theta_1/2))^2+(sin (theta_2/2)-cos(theta_2/2))^2 +(sin (theta_3/2)-cos(theta_3/2))^2=0#
Sum of three squared quantities being zero each quantity should be zero
So
#sin (theta_1/2)-cos(theta_1/2)=0#
#=>tan (theta_1/2)=1=tan(pi/4)#
Hence #theta_1=pi/2#
Similarly
#theta_2=pi/2#
and
#theta_3=pi/2#
then the value of #cos theta_1+ cos theta_2 + cos theta _3#
#=cos (pi/2)+ cos (pi/2) + cos(pi/2)=0+0+0=0#