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 $0<=x<=2pi$

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