Question

The sum of roots of ${\sin }^{2}x-5\sin x\cos x+2=0$, where $x\in [0,2\pi ]$ is

• A. $\frac{3\pi }{2}$
• B. $\frac{5\pi }{2}$
• C. $\frac{5\pi }{2}+{\tan }^{-1}\frac{2}{3}$
• D. $\frac{5\pi }{2}+{\tan }^{-1}\left(\frac{12}{5}\right)$

Answer 1

The correct option is D $\frac{5\pi }{2}+{\tan }^{-1}\left(\frac{12}{5}\right)$

${\sin }^{2}x-5\sin x\cos x+2=0$
As $\cos x\ne 0$, dividing by ${\cos }^{2}x$, we get
$\Rightarrow {\tan }^{2}x-5\tan x+2{\sec }^{2}x=0\Rightarrow 3{\tan }^{2}x-5\tan x+2=0\Rightarrow 3{\tan }^{2}x-3\tan x-2\tan x+2=0\Rightarrow (3\tan x-2)(\tan x-1)=0\Rightarrow \tan x=1,\frac{2}{3}$
As $x\in [0,2\pi ]$, so
$\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},{\tan }^{-1}\frac{2}{3},\pi +{\tan }^{-1}\frac{2}{3}$
Hence, the sum of roots
$=\frac{5\pi }{2}+2{\tan }^{-1}\left(\frac{2}{3}\right)=\frac{5\pi }{2}+{\tan }^{-1}\frac{\frac{4}{3}}{1-\frac{4}{9}}=\frac{5\pi }{2}+{\tan }^{-1}\left(\frac{12}{5}\right)$