If 0≤x≤π and 81sin2⁡x+81cos2⁡x=30, then x is equal to

Let 81sin2⁡x=y. Then, 81cos2⁡x=811−sin2⁡x=81y−1Now, 81sin2⁡x+81cos2⁡x=30⇒y+81y=30⇒y2−30y+81=0⇒ y=3 or, y=27⇒ 81sin2⁡x=3 or  81sin2⁡x=27=34sin2⁡x=31 or 34sin2⁡x=33⇒ 4sin2⁡x=1,4sin2⁡x=3⇒ sin⁡x=±12 or, sin⁡x=±32⇒x=π6 or π3