Question

The general solution of $1+{\mathrm{Sin}}^{2}x=3\mathrm{Sin}x\cdot \mathrm{Cos}x,\mathrm{Tan}x\ne \frac{1}{2}$ is ......

• A. $n\pi -\frac{\pi }{4},n\in Z$
• B. $n\pi +\frac{\pi }{4},n\in Z$
• C. $2n\pi +\frac{\pi }{4},n\in Z$
• D. $2n\pi -\frac{\pi }{4},n\in Z$

Answer 1

Answer: (B)

$1+{\sin }^{2}x=3\sin x\cdot \cos x,\tan x\ne \frac{1}{2}$
Divided by ${\cos }^{2}x$ on both sides,
$\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}=3\frac{\sin x\cdot \cos x}{\cos x\cdot \cos x}$
${\sec }^{2}x+{\tan }^{2}x=3\tan x$
$1+{\tan }^{2}x+{\tan }^{2}x=3\tan x$
$2{\tan }^{2}x-3\tan x+1=0$
$2{\tan }^{2}x-2\tan x-\tan x+1=0$
$2\tan x(\tan x-1)-1(\tan x-1)=0$
$(\tan x-1)(2\tan x-1)=0$
$\tan x=1,\frac{1}{2}$
We take, $\tan x=1\left(\because \tan x\ne \frac{1}{2}\right)$
$\tan x=\tan (\pi /4)$
$x=n\pi +\pi /4,n\in Z$