Question

If $cos2x=3cosx+1=\frac{cosexx}{cotx-cot2x}$, then which of the following is true ?

• A. $x=(2n+1)\frac{\pi }{2},n\epsilon I$
• B. $x=2n\pi ,n\epsilon I$
• C. $x=2n\pi \pm co{s}^{-1}\left(\frac{2}{5}\right),n\epsilon I$
• D. no real x

Answer 1

The correct option is D no real x

$\underset{a}{\underset{\downarrow }{\cos 2x}}=\underset{b}{\underset{\downarrow }{3\cos x+1}}=\underset{c}{\underset{\downarrow }{\frac{cosecx}{\cot x-\cot 2x}}}$

(i) $a=b$

$\Rightarrow 2{\cos }^{2}x-1=3\cos x+1$

$\Rightarrow 2{\cos }^{2}x-3\cos x-2=0$

$\Rightarrow 2{\cos }^{2}x-4\cos x+\cos x-2=0$

$\Rightarrow 2\cos x[\cos x-2]+1[\cos x-2]=0$

$\Rightarrow \underset{\downarrow }{\cos x=2}$ or $\cos x=\frac{-1}{2}$

not possible $x\in \left[\right(2x+1)\pi -\frac{\pi }{3}]$ .........(1)

(ii) $b=c$

$\frac{\left(\frac{1}{\sin x}\right)}{(\frac{\cos x}{\sin x}-\frac{\cos 2x}{\sin x})}=3\cos +1$

$\Rightarrow \frac{\frac{1}{\sin x}}{\left[\frac{\sin x\cos x-\cos x\sin x}{\sin x\sin x}\right]}=3\cos x+1$

$\Rightarrow \frac{\frac{1}{\sin x}}{\left[\frac{\sin (2x-x)}{\sin x\sin 2x}\right]}=3\cos +1$

$\Rightarrow \frac{2\sin x-\cos x}{\sin x}=3\cos x+1$

$\Rightarrow \cos x=-1$

$\Rightarrow x=(2n+1)\pi$

from (1) and (2)

these no $x$ thatr satisfies both condition

$\therefore$ no real $x$ exits.