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

Options

A. 2
B. 4
C. 6
D. 8

Answer

D

Explanation

We have,
`16^("sin"^(2)x) + 16^("cos"^(2) x) = 10`
`rArr 16^("sin"^(2)x) + (16)/(16^("sin"^(2)x)) = 10`
`rArr y^(2) - 10y + 16 =0, " where" y= 16^("sin"^(2)x)`
`rArr y = 2, 8`
`rArr 16^("sin"^(2)x) = 2, 8`
`rArr 2^(4"sin"^(2)x) = 2^(1), 2^(3)`
`rArr 4"sin"^(2) x = 1, 3`
`rArr "sin"^(2)x = (1)/(4), (3)/(4)`
`rArr "sin" x = +- (1)/(2) , +- (sqrt(3))/(2)`
`rArr x = (pi)/(6), (5pi)/(6), (7pi)/(6), (11pi)/(6), (pi)/(3), (2pi)/(3), (4pi)/(3), (5pi)/(3)`
Thus, there are 8 solutions.