The value of x satisfying the equation and is

Engineering

Mathematics

Introduction to Determinants

Question

The value of x satisfying the equation $|\begin{array}{lll} \cos 2 x \sin 2 x \sin 2 x \\ \sin 2 x \cos 2 x \sin 2 x \\ \sin 2 x \sin 2 x \cos 2 x \end{array}| = 0$ and $x ε [0, \frac{π}{4}]$ is

$\frac{π}{2}$

$\frac{π}{16}$

$\frac{π}{3}$

$\frac{π}{8}$

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Given, $|\begin{array}{lll} \cos 2 x \sin 2 x \sin 2 x \\ \sin 2 x \cos 2 x \sin 2 x \\ \sin 2 x \sin 2 x \cos 2 x \end{array}| = 0; x ε [0, \frac{π}{4}]$

Applying operation : R₁→ R₁ + R₂ + R₃, we get
$|\begin{array}{ccc} \cos 2 x + 2 \sin 2 x & 2 \sin 2 x + \cos 2 x & 2 \sin 2 x + \cos 2 x \\ \sin 2 x & \cos 2 x & \sin 2 x \\ \sin 2 x & \sin 2 x & \cos 2 x \end{array}| = 0$

$\Rightarrow (2 \sin 2 x + \cos 2 x) |\begin{array}{ccc} 1 & 1 & 1 \\ \sin 2 x & \cos 2 x & \sin 2 x \\ \sin 2 x & \sin 2 x & \cos 2 x \end{array}| = 0$
Now, apply operation
C₂ → C₂ – C₁ and C₃ → C₃ – C₁, we get
(2sin2x + cos2x)
$|\begin{array}{ccc} 1 & 0 & 0 \\ \sin 2 x & \cos 2 x - \sin 2 x & 0 \\ \sin 2 x & 0 & \cos 2 x - \sin 2 x \end{array}| = 0$
Expanding along R₁, we get
(2sin2x + cos2x)(cos2x – sin2x)² = 0
⇒ 2sin2x + cos2x = 0
or cos2x – sin2x = 0
$\Rightarrow \tan 2 x = - \frac{1}{2}$
or $\tan 2 x = 1 = \tan \frac{π}{4}$
$∵ x \neq \frac{1}{2} \tan^{- 1} (\frac{- 1}{2})$
$∴ x = \frac{π}{8} ε [0, \frac{π}{4}] .$

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