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JEE Advanced · Mathematics · 8. Trigonometric Equations

The number of all possible values of \(\theta\), where \(0 < \theta < \pi\), for which the system of equations
\((y+z) \cos 3 \theta=(x y z) \sin 3 \theta \) \( x \sin 3 \theta=\frac{2 \cos 3 \theta}{y}+\frac{2 \sin 3 \theta}{z}\) 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

  1. A 1
  2. B 6
  3. C 7
  4. D 3
Verified Solution

Answer & Solution

Correct Answer

(D) 3

Step-by-step Solution

Detailed explanation

Given equations can be written as
\(x \sin 3 \theta-\frac{\cos 3 \theta}{y}-\frac{\cos 3 \theta}{z}=0 \)
\( x \sin 3 \theta-\frac{2 \cos 3 \theta}{y}-\frac{2 \sin 3 \theta}{z}=0 \)
\( \text { and } x \sin 3 \theta-\frac{2}{y} \cos 3 \theta \)
\( -\frac{1}{z}(\cos 3 \theta+\sin 3 \theta)=0\)
Eqs. (ii) and (iii), implies
\(2 \sin 3 \theta=\cos 3 \theta+\sin 3 \theta \)
\( \Rightarrow \sin 3 \theta=\cos 3 \theta \)
\( \therefore \tan 3 \theta=1 \)
\( \Rightarrow 3 \theta=\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4} \text { or } \theta=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{9 \pi}{12}.\)