`"If cos" 2 theta, 1 "and sin" theta "are in GP., then "theta =`

Options

A. `npi+ (-1)^(2) (pi)/(2), n in Z`
B. `n pi + (-1)^(n-1) (pi)/(2), n in Z`
C. `2n pi, n in Z`
D. none of these

Answer

B

Explanation

It is given that `"cos" 2 theta, 1 "and sin" theta` are in G.P.
therefore,
`1 = "sin" theta "cos" 2 theta`
`rArr "sin" theta (1-2 "sin"^(2) theta) = 1`
`rArr 2 "sin"^(3) theta - "sin" theta + 1 =0`
`rArr ("sin" theta + 1)(2 "sin"^(2) theta - 2 "sin" theta + 1) = 0`
`rArr "sin" theta + 1 =0 " " [because 2 "sin"^(2) theta - 2"sin" theta + 1 gt 0 " for all "theta]`
`rArr "sin"theta = "sin" (-(pi)/(2))`
`rArr theta = n pi + (-1)^(n) (-(pi)/(2)), n in Z`
` theta = n pi + (-1)^(n-1) (pi)/(2), n in Z`