Solve: `16^sin^(2x)16^cos^(2x)=10 ,0lt=x<2pi`

Explanation

`16^(sin^(2)x)+16^(1-sin^(2) x)=10`
If `16^(sin^(2)x)=t`, then `t+16/t=10`
Then Eq. (i) becomes
`t^(2)-10t+16=0`
or `t=2, 8`
`rArr 16^(sin^(2) x)=16^(1//4)` or `16^(3//4)`
`rArr sin x= pm 1/2, pm sqrt(3)/2`
Now `sin x=1/2`, then `x=pi/6, (5pi)/6`
`sin x=-1/2`, then `x=(7pi)/6` or `(11pi)/6`
`sin x= sqrt(3)/2`, then `x=pi/3, (2pi)/3`
`sin x= - sqrt(3)/2`, then `x=(4pi)/3, (5 pi)/3`
Hence, there will be eight solutions in all.