Question

List 1 contains different equations while List 2 contains solutions. Match options from List 1 with options from List 2

Answer 1

A) $7{\cos }^{2}x+\sin x\cos x-3=0$
$\Rightarrow 3{\sin }^{2}x-\sin x\cos x-4{\cos }^{2}x=0\Rightarrow 3{\tan }^{2}x-\tan x-4=0$
$\Rightarrow (3\tan x-4)(\tan x+1)=0\Rightarrow \tan x=\frac{4}{3}$ or $\tan x=-1$

B) $(2{\sin }^2\left(\frac{\pi }{2}{\cos }^{2}x\right)=1-\cos \left(\pi \sin 2x\right)$
$\Rightarrow (2{\sin }^2\left(\frac{\pi }{2}{\cos }^{2}x\right)=2{\sin }^2\left(\frac{\pi }{2}\sin 2x\right)\Rightarrow {\cos }^{2}x=\sin 2x\Rightarrow \cos x(\cos x-2\sin x)=0$
$\Rightarrow \cos x=0$ or $\tan x=\frac{1}{2}\Rightarrow \cos 2x=\frac{3}{5}$

C) $6se{c}^{2}x-11\tan x-2=0\Rightarrow 6{\tan }^{2}x-11\tan x+4=0$
$\Rightarrow (3\tan x-4)(2\tan x-1)=0\Rightarrow \tan x=\frac{4}{3}$ or $\tan x=\frac{1}{2}\Rightarrow \cos 2x=\frac{3}{5}$

D) $2\cos 2x-\sin 2x=2{\sin }^{2}c\Rightarrow 1-2{\sin }^{2}x-\sin x\cos x={\sin }^{2}x$
$\Rightarrow 3{\sin }^{2}x+\sin x\cos x-1=0\Rightarrow 2{\sin }^{2}x+\sin \cos x-{\cos }^{2}x=0\Rightarrow 2{\tan }^{2}x+\tan x-1=0\Rightarrow (2\tan x-1)(\tan x+1)=0$
$\Rightarrow \tan x=-1$ or $\tan x=\frac{1}{2}\Rightarrow \cos 2x=\frac{3}{5}$