81cos2x+81sin2x=30
⇒81sin2x+811−sin2x=30
⇒81sin2x+8181sin2x=30
Let 81sin2x=t
Then t+81t=30
⇒t2−30t+81=0
⇒(t−3)(t−27)=0
⇒t=3,27
81sin2x=3 and 81sin2x=27
⇒34sin2x=31 and 34sin2x=33
⇒sin2x=14 and sin2x=34
⇒sinx=±12 and sinx=±√32
Since x∈[0,2π],
∴x=π6,5π6,7π6,11π6,π3,2π3,4π3,5π3