The number of all possible values of where 0 , for which the system of equations ( y + z ) cos 3 = ( x y z ) sin 3 x sin 3 = 2 cos 3 y + 2 sin 3 z ( x y z ) sin 3 = ( y + 2 z ) cos 3 + y sin 3 have a solution ( x 0 , y 0 , z 0 ) with y 0 z 0

Question

The number of all possible values of $θ$ where $0 < θ < π,$ for which the system of equations

$(y + z) \cos 3 θ = (x y z) \sin 3 θ$

$x \sin 3 θ = \frac{2 \cos 3 θ}{y} + \frac{2 \sin 3 θ}{z}$

$(x y z) \sin 3 θ = (y + 2 z) \cos 3 θ + y \sin 3 θ$

$have a solution (x_{0}, y_{0}, z_{0}) with y_{0} z_{0} \neq 0, is.$ [2010]

Smart Achievers · Accepted Answer

(3)

Given equations are

$x y z \sin 3 θ = (y + z) \cos 3 θ . . . (i)$

$x y z \sin 3 θ = 2 z \cos 3 θ + 2 y \sin 3 θ . . . (i i)$

$x y z \sin 3 θ = (y + 2 z) \cos 3 θ + y \sin 3 θ . . . (i i i)$

$On subtracting eq. (ii) from (i), we get$

$(\cos 3 θ - 2 \sin 3 θ) y - (\cos 3 θ) z = 0 . . . (i v)$

$On subtracting eq. (i) from (iii), we get$

$\sin 3 θ y + (\cos 3 θ) z = 0 . . . (v)$

$Eq. (iv) and (v) form a homogeneous system of linear equations.$

$But y \neq 0, z \neq 0$

$∴$ $\frac{\cos 3 θ - 2 \sin 3 θ}{\sin 3 θ} = - \frac{\cos 3 θ}{\cos 3 θ} \Rightarrow \cos 3 θ = \sin 3 θ$

$\Rightarrow \tan 3 θ = 1 \Rightarrow 3 θ = n π + \frac{π}{4}$ $\Rightarrow θ = (4 n + 1) \frac{π}{12}, n \in ℤ$

$For θ \in (0, π) \Rightarrow θ = \frac{π}{12}, \frac{5 π}{12}, \frac{3 π}{4}$

$∴$ $Three such solutions are possible.$

Want Us to Email you About Special Offers & Updates?