The number of all possible values of `theta`, where `0 lt theta lt pi`, for which the system of equations `(y+z)cos 3 theta =(xyz) sin 3 theta ,x sin 3 theta =(2cos3theta)/y+(2sin3theta)/z and (x y z)sin3theta=(y+2z)cos3theta+ysin3theta` have a solution `(x_0,y_0,z_0)` wiith `y_0 z_0 !=0` is

Options

A. 0
B. 2
C. 3
D. 4

Answer

C

Explanation

We have,
`(y +z) "cos" 3theta = xyz "sin" 3theta " "..(i)`
`x "sin" 3theta = (2"cos" 3theta)/(y) + (2"sin" 3theta)/(z) " "… (ii)`
`xyz "sin" 3theta = (y +2z)"cos" 3theta + y"sin"3theta " "....(iii)`
From (i) and (ii), we get
`y ("cos" 3theta-2"sin" 3theta)-z"cos" 3theta = 0 " "...(iv)`
From (ii) and (iii), we get
`y("cos" 3theta- "sin" 3theta) = 0" "...(v)`
Since the given system of equations have a solutoin `(x_(0), y_(0), z_(0))` such that `y_(0)z_(0) ne 0`
`therefore "cos" 3theta -"sin" 3theta = 0 " "["From"(v)]`
`rArr "tan" 3theta = 1`
`rArr 3theta = (pi)/(4), (5pi)/(4), (9pi)/(4) " " [because 0 lt theta lt pi]`
`rArr theta = (pi)/(12), (5pi)/(12), (3pi)/(4)`
Hence, theta are three values of `theta`.