The number of all possible values of θ , where 0 θ π , for which the system of equations array r (y+z) 3 θ=(x y z) 3 θ x 3 θ=(2 3 θ)/(y)+(2 3 θ)/(z) array and (x y z) 3 θ=(y+2 z) 3 θ+y 3 θ have a solution (x_0, y_0

Correct answer: 3 Given equations can be written as gathered x 3 θ-( 3 θ)/(y)-( 3 θ)/(z)=0 x 3 θ-(2 3 θ)/(y)-(2 3 θ)/(z)=0 and x 3θ-(2)/(y) 3 θ -(1)/(z)( 3 θ+ 3 θ)=0 gathered Eqs. (ii) and (iii), implies aligned & 2 3 θ= 3 θ+ 3 θ & 3 θ= 3 θ & 3 θ=1 & 3 θ=(π)/(4), (5 π)/(4), (9 π)/(4) or θ=(π)/(12), (5 π)/(12), (9 π)/(12) . aligned

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Trigonometric Equations

JEE Advanced 2010 · Paper 1 · Mathematics · Trigonometric Equations

The number of all possible values of θ , where 0 θ π , for which...

The number of all possible values of

\theta

, where

0 \lt \theta \lt \pi

, for which the system of equations

\begin{array}{r} (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} \end{array}

and

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

have a solution

\left(x_0, y_0, z_0\right)

with

\quad y_0 z_0 \neq 0

, is

Correct answer

Step-by-step solution

Given equations can be written as

\begin{gathered}x \sin 3 \theta-\frac{\cos 3 \theta}{y}-\frac{\cos 3 \theta}{z}=0 \\ x \sin 3 \theta-\frac{2 \cos 3 \theta}{y}-\frac{2 \sin 3 \theta}{z}=0 \\ \text{ and }x \sin 3\theta-\frac{2}{y} \cos3 \theta\\ -\frac{1}{z}(\cos3 \theta+\sin3 \theta)=0 \end{gathered} Eqs. (ii) and (iii), implies

\begin{aligned}& 2 \sin 3 \theta=\cos3 \theta+\sin3 \theta\\ \Rightarrow & \sin 3 \theta=\cos3 \theta\\ \therefore & \tan 3 \theta=1 \\ \Rightarrow & 3 \theta=\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4} \text{ or } \theta=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{9 \pi}{12} . \end{aligned}

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