Question

Common roots of the equations $2si{n}^{2}x+si{n}^{2}2x=2$ and $sin2x+cos2x=tanx,$ are

• A. $x=(2n-1)\frac{\pi }{2}$
• B. $x=(2n+1)\frac{\pi }{4}$
• C. $x=(2n+1)\frac{\pi }{3}$
• D. None of these

Answer 1

The correct option is D None of these

$2{\sin }^{2}x+{\sin }^{2}2x=2$

$\Rightarrow 2{\sin }^{2}x+{\left(2\sin x\cos x\right)}^2=2$

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

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

$\Rightarrow 2{\sin }^{2}x+4{\sin }^{2}x(1-{\sin }^{2}x)-2=0$

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

$\Rightarrow -4{\sin }^{4}x+6{\sin }^{2}x-2=0$

$\Rightarrow 2{\sin }^{4}x-3{\sin }^{2}x+1=0$

$\Rightarrow 2{\sin }^{4}x-2{\sin }^{2}x-{\sin }^{2}x+1=0$

$\Rightarrow 2{\sin }^{2}x({\sin }^{2}x-1)-({\sin }^{2}x-1)=0$

$\Rightarrow ({\sin }^{2}x-1)(2{\sin }^{2}x-1)=0$

$\Rightarrow {\sin }^{2}x=1$ or $2{\sin }^{2}x=1$

$\Rightarrow {\sin }^{2}x=1$ or ${\sin }^{2}x=\frac{1}{2}$

$\Rightarrow \sin x=\pm 1$ or $\sin x=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow x=(2n+1)\frac{\pi }{2}$ or $x=n\pi \pm \frac{\pi }{4}$

....$\left(1\right)$

$\sin 2x+\cos 2x=\tan x$

$\Rightarrow \frac{2\tan x}{1+{\tan }^{2}x}+\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=\tan x$

$\Rightarrow \frac{2\tan x+1-{\tan }^{2}x}{1+{\tan }^{2}x}=\tan x$

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

$\Rightarrow \tan x+{\tan }^{2}x-2\tan x-1+{\tan }^{3}x=0$

$\Rightarrow {\tan }^{3}x+{\tan }^{2}x-\tan x-1=0$

$\Rightarrow {\tan }^{2}x(\tan x+1)-(\tan x+1)=0$

$\Rightarrow (\tan x+1)({\tan }^{2}x-1)=0$

$\Rightarrow \tan x=1,or{\tan }^{2}x=1$

$\Rightarrow x=n\pi \pm \frac{\pi }{4}$ ....$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ Common roots are $n\pi \pm \frac{\pi }{4}$