√3sinx+2cosx−cos2x−1=0⇔√3sinx+2cosx−cos2x+sin2x−1=0
⇔√3sinx+sin2x−(cos2x−2cosx+1)=0
⇔sinx(sinx+√3)−(1−cosx)2=0
⇔2sinx2cosx2(sinx+√3)−2sin2x2(1−cosx)=0
⇔sinx2=0∨cosx2(sinx+√3)−sinx2(1−cosx)=0
∗sinx2=0⇔x=k2π(k∈Z)
∗cosx2(sinx+√3)−sinx2(1−cosx)=0
⇔sinxcosx2+cosxsinx2−sinx2+√3cosx2=0
⇔sin3x2−2sin(x2−π3)=0(2)
Đặt t=x2−π3⇒3x2=3t+π⇒sin3x2=sin(3t+π)=−sin3t
(2)⇔−sin3t−2sint=0⇔4sin3t−5sint=0
⇔sin2t=54(VL)∨sint=0⇔t=x2−π3=hπ⇔x=2π3+h2π(h∈Z)