2cos2x−1+1−√3=(2−√3)cosx
⇒cosx(cosx−1)+√3(cosx−1)=0
⇒(cosx−1)(2cosx+√3)=0
⇒cosx=1 or cosx=−√32
sin3x=2sinx
⇒3sinx−4sin3x=2sinx
⇒sinx=0 or 3−4sin2x=2
4sin2x=1
⇒sinx=±12
Common solutions in [0,5π] are 0,2π,4π,5π6,7π6,17π6,19π6,29π6
If tanx≥1√3
⇒x∈[π6+nπ,π2+nπ];n∈Z
So, common solutions x=7π6,19π6