The number of all possible values of \(\theta\), where \(0<\theta<\pi\), for which the system of equations \[ \begin{gathered} (y+z) \cos 3 \theta=(x y z) \sin 3 \theta \\ x \sin 3 \theta=\frac{2 \cos 3 \theta}{y}+\frac{2 \sin 3 \theta}{z} \\ (x y z) \sin 3 \theta=(y+2 z) \cos 3 \theta+y \sin 3 \theta \end{gathered} \] have a solution \(\left(\mathrm{x}_{0}, \mathrm{y}_{0}, \mathrm{z}_{0}\right)\) with \(\mathrm{y}_{0} \mathrm{z}_{0} \neq 0\), is

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Trigonometric Equations

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

Answer: 3

From the first equation:

(y + z) cos 3 θ = (x y z) sin 3 θ

From the second equation:

x y z sin 3 θ = 2 z cos 3 θ + 2 y sin 3 θ

Substituting into the first equation:

(y + z) cos 3 θ = 2 z cos 3 θ + 2 y sin 3 θ

From the third equation:

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

Comparing these equations:

y (cos 3 θ - 2 sin 3 θ) = z cos 3 θ

y (sin 3 θ - cos 3 θ) = 0

Since

y_{0} \neq = 0

, we must have:

sin 3 θ = cos 3 θ

tan 3 θ = 1

3 θ = nπ + \frac{π}{4}

For

0 < θ < π

, we have

0 < 3 θ < 3 π

.

The solutions are:

3 θ = \frac{π}{4}, \frac{5 π}{4}, \frac{9 π}{4}

This gives

θ = \frac{π}{12}, \frac{5 π}{12}, \frac{3 π}{4}

Therefore, there are 3 values.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer