Can anyone help me to solve# 4sin(x-pi)cos(x-pi)=√3 " for " 0<x<2pi#?

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Answer 1 · 2017-07-21T13:43:01

# 4sin(x-pi)cos(x-pi)=sqrt3#

# =(-4sin(x))(-cos(x))=sqrt3#

# =4sin(x)cos(x)=sqrt3#

# =2*sin(2x)=sqrt3#

# =sin(2x)=sqrt3/2=sin(pi/3)=sin(pi-pi/3)=sin(2pi+pi/3)=sin(3pi-pi/3)#

So

#2x=pi/3#

#=>x=pi/6#

And

#2x=(2pi)/3#

#=>x=pi/3#

Again

#sin(2x)=sin(2pi+pi/3)#

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

Lastly

#sin(2x)=sin(3pi-pi/3)#

#=>x=(8pi)/6=(4pi)/3#