Question #7f803

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Answer 1 · 2017-01-30T05:13:45

#4sinxcosx-2sqrt3sinx+2cosx-sqrt3=0#

#=>2sinx(2cosx-sqrt3)+1(2cosx-sqrt3)=0#

#=>(2cosx-sqrt3)(2sinx+1)=0#

So #2cosx-sqrt3=0#
#=>cosx=sqrt3/2=cos(pi/6)=cos(2pi-pi/6)=cos((11pi)/6)#

Hence

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

Again

#2sinx+1=0#

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

So #=>x=(7pi)/6#

Again

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

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

Hence solutions for #0<=x<=2pi#

#x = pi/6,(7pi)/6,(11pi)/6#