16sin2x+16cos2x=10
or 16sin2x+161−sin2x=10
or 16sin2x+16.16−sin2x=10
put 16sin2x=t
Then t+16t=10ort2−10t+16=0
or (t−2)(t−8)=0 , giving t=2or8
Hence 16sin2x=2 or 24sin2x=2, which gives
4sin2x=1
or sin2x=14=sin2(π6).....(1)
∴ solution of (1) is, x=nπ±π6
where n=0,±1,±2,±3 , .......
Again t=8gives24sin2x=8=23
or 4sin2x=3i.e.sin2x=34=sin2π3....(2)
∴ The solution of (2) is , x=nπ±π3
where n=0,±1,±2,±3.......