Question

If sum of all $x$ in $[0,2\pi ]$ such that $3{\cot }^{2}x+8\cot x+3=0$ is $K\pi$ then find $K$.

Answer 1

$3{\cot }^{2}x+8\cot x+3=0$

Let $\cot x=y$

$3y^2+8y+3=0$

Let the roots be $y_2$ and $y_2$

Sum of roots : $y_1+y_2=\frac{-8}{3}$

Product of roots: $4y_1{y}_2=\frac{3}{3}=1$

$y_1=\frac{1}{y_2}$

Since, $y=\cot x$

$\therefore \cot x_1=\frac{1}{\cot x_2}$

$\cot x_1=\tan x_2$

$\tan x_2=\tan (\frac{3z}{2}-x_1)$

So,

$x_2=\frac{3z}{2}-x_1$

$x_2+x_1=\frac{3z}{2}$

Value of $K=\frac{3}{2}$