If 0≤x≤π and 81sin2x+81cos2x=30, then x is equal to
π6,π3
π3,π2
5π6,π3
2π3,π3
Let 81sin2x=y. Then, 81cos2x=811−sin2x=81y−1
Now,
81sin2x+81cos2x=30⇒y+81y=30⇒y2−30y+81=0
⇒ y=3 or, y=27
⇒ 81sin2x=3 or 81sin2x=27
=34sin2x=31 or 34sin2x=33
⇒ 4sin2x=1,4sin2x=3
⇒ sinx=±12 or, sinx=±32⇒x=π6 or π3