This is a trigonometric equation I can't figure out? cot^2x = -2cotx -1

Apr 27, 2018

$x = \left(2 k + 1\right) \pi - \frac{\pi}{4} , k \in \mathbb{Z}$

Explanation:

Here,

$C o {t}^{2} x = - 2 \cot x - 1$

$\implies {\cot}^{2} x + 2 \cot x + 1 = 0$

$\implies {\left(\cot x + 1\right)}^{2} = 0$

$\implies \cot x + 1 = 0$

color(red)(=>cotx=-1...(A)

$\implies \tan x = - 1. . .$ $\to \cos x \ne 0$

$\implies \tan x = \tan \left(- \frac{\pi}{4}\right)$

color(blue)(=>x=kpi-pi/4,kinZZ...to(B)

We know that ,the range of ${\cot}^{-} 1 x , i s : \left(0. \pi\right)$

Now from $\left(A\right)$

$\cot x = - 1 \implies x = {\cot}^{-} 1 \left(- 1\right) \ne - \frac{\pi}{4.} . . \to I {V}^{t h} Q u a \mathrm{dr} a n t$

$\mathmr{and} - \frac{\pi}{4} \notin \left(0 , \pi\right)$

So, color(red)(cot^-1(-1)=pi-pi/4=(3pi)/4...toII^(nd)Quadrant

Hence, from $\left(B\right)$

$x = \left(2 k + 1\right) \pi - \frac{\pi}{4} , k \in \mathbb{Z}$

Note:

color(blue)(x={color(red)(kpi)-pi/4,kinZZ}

:.x={color(red)((2k+1)pi)-pi/4,kinZZ}uu{color(red)((2k)pi)- pi/4,kinZZ}
color(white)(.................)color(red)(II^(nd)Quadrant)color(white) (...................)color(red)(IV^(th)Quadrant)