Question

$\cos \theta \cos 2\theta \cos 3\theta =\frac{1}{4}(0\le \theta \le \pi )$
If $\alpha$ is a root of this equation, $2\cos \alpha$ is a root of the equation

• A. $x^2-1=0$
• B. $x^2+1=0$
• C. $x^4-4x^2+2=0$
• D. $x^4-4x^2+4=0$

Answer 1

Answer: (A), (C)

The correct options are
A $x^2-1=0$
C $x^4-4x^2+2=0$

$\cos \theta \cos 2\theta \cos 3\theta =\frac{1}{4}(0\le \theta \le \pi )$
$\Rightarrow 2\cos 2\theta (\cos 4\theta +\cos 2\theta )=1$
$\Rightarrow 2\cos 2\theta \cos 4\theta +2{\cos }^{2}2\theta =1$
$\Rightarrow \cos 4\theta (2\cos 2\theta +1)=0$
$\Rightarrow$ Either $\cos 4\theta =0,or\cos 2\theta =-\frac{1}{2}$.
$\cos 4\theta =0\Rightarrow \theta =(2n+1)\pi /8n\epsilon I.$
$\Rightarrow \theta =\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8}$ as$0\le \theta \le \pi$ and
$\cos 2\theta =-\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}$ and $\frac{2\pi }{3}$ again as $0\le \theta \le \pi$
Next, let $x=2\cos \alpha$, where $\alpha$ is a root of the equation, i.e., either $\cos 4\alpha =0$ or $\cos 2\alpha =-\frac{1}{2}$.
So, $x^2=4{\cos }^2\alpha =2(1+\cos 2\alpha )=1$
If $\cos 2\alpha =-\frac{1}{2}$ $\Rightarrow 2\cos \alpha$ is a root of the equation $x^2-1=0$, or
$x^2-2=2\cos 2\alpha$ $\Rightarrow (x^2-2{)}^2=4{\cos }^{2}2\alpha =2(1+\cos 4\alpha )=2$, if $\cos 4\alpha =0$$\Rightarrow x^4-4x^2+2=0\Rightarrow 2\cos \alpha$ is a root of the equation $x^4-4x^2+2=0$.
Hence, options A and C are the correct answers.