Đặt
x=t+π6pt đã cho ⇔4sin(t+π3).(sin(2t+π2)−1)=2cos(2t+π3)−1
⇔4sin(t+π3)(cos2t−1)=2(cos2t.12−sin2t.√32)−1
⇔4sin(t+π3)(cos2t−1)=(2cos2t−1)−√3sin2t
⇔[4sin(t+π3)−1](cos2t−1)=−√3sin2t
⇔[4sin(t+π3)−1](−2sin2t)=−2√3sintcost
⇔[sint=0(1)−2sint(2sint+2√3cost−1)=−2√3cost(2)
(2)⇔−2sint(2sint−1)=−2√3cost+4√3sintcost
⇔−2sint(2sint−1)=2√3cost(2sint−1)
⇔[2sint=1(3)−sint=√3cost(4)
Giải (1)(3)(4) tìm được nghiệm