Question

How do you solve `2cosx+1=0` for `0°<=x<=360°`?

Answer 1

`x=120^@" or "x=240^@`

Explanation:

`2cosx+1=0`

`"subtract 1 and divide by 2"`

`2cosx=-1rArrcosx=-1/2`

`"since "cosx<0" then x in second/third quadrants"`

`x=cos^-1(1/2)=60^@larr" related acute angle"`

`x=(180-60)^@=120^@larrcolor(red)"second quadrant"`

`x=(180+60)^@=240^@larrcolor(red)"third quadrant"`

`x =120^@,240^@to0<=x<=360`

Answer 2

`x=120^circ or240^circ`

Explanation:

Here,

`2cosx+1=0 , where, 0^circ <= x <= 360^circ`

`=>2cosx=-1`

`=>cosx=-1/2 < 0=>II^(nd)Quadrant or III^(rd)Quadrant`

`(i)90^circ < x < 180^circ=>x=120^circto(II^(nd)Quadrant)`

`(ii)180^circ < x < 270^circ=>x=240^circto(III^(rd)Quadrant)`

Hence,

`x=120^circ or240^circ`