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Answer 1 · 2017-02-10T05:58:10

#x=(7pi)/6 and (11pi)/6" when "x in [0,2pi]#

#2sin^2x-9sinx=5#
#=>2sin^2x-9sinx-5=0#

#=>2sin^2x-10sinx+sinx-5=0#

#=>2sinx(sinx-5)+1(sinx-5)=0#

#=>(sinx-5)(2sinx+1)=0#

#sinx=5->"not possible"#

#2sinx+1=0#

#=>sinx=-1/2=-sin(pi/6)#

#=>sinx=sin(pi+pi/6)=sin((7pi)/6)#
So #x=(7pi)/6#

Again

#=>sinx=-1/2=-sin(pi/6)#

#=>sinx=sin(2pi-pi/6)=sin((11pi)/6)#

#=>x=(11pi)/6#

Hence solutions are

#x=(7pi)/6 and (11pi)/6" when "x in [0,2pi]#