2cos2x−1+1−√3=(2−√3)cos x
2cosx(cosx−1)+√3(cosx−1)=0
(cosx−1)(2cosx+√3)=0
cos x=1 or cos x=−√32
sin 3x=2 sin x
⇒3 sin x−4 sin3 x=2 sin x
sin x=0 or 3−4sin2x=2
4sin2x=1
sin x=±12
common solution in [0,5π] is 0,2π,4π,5π6,7π6,17π6,19π6,29π6
If tan x≥1√3⇒x∈[π6+nπ,π2+nπ];n∈I
So, common solution:x=7π6,19π6