Question

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

Answer 1

`x=(2k+1)pi-pi/4,kinZZ`

Explanation:

Here,

`Cot^2x = -2cotx -1`

`=>cot^2x+2cotx+1=0`

`=>(cotx+1)^2=0`

`=>cotx+1=0`

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

`=>tanx=-1...` `tocosx!=0`

`=>tanx=tan(-pi/4)`

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

We know that ,the range of `cot^-1x, is :(0.pi)`

Now from `(A)`

`cotx=-1=>x=cot^-1(-1)!=-pi/4...toIV^(th)Quadrant`

`and-pi/4!in(0,pi)`

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

Hence, from `(B)`

`x=(2k+1)pi-pi/4,kinZZ`

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)`