Question

How do you solve `2cosx+4=5`?

Answer 1

`x=pi/3+2kpi" or "x=(5pi)/3+2kpi`

Explanation:

`2cosx+4=5`

`rArr2cosx=5-4=1`

`rArrcosx=1/2`

`rArrx=cos^-1(1/2)=pi/3`

`"since "cosx>0" then x is in first/fourth quadrants"`

`rArrx=2pi-pi/3=(5pi)/3`

`"because cos is periodic it will have solutions at"`
`"multiples of "2pi`

`"thus we can express the solutions in general terms as"`

`x=pi/3+2kpitok inZZ`

`k=(5pi)/3tok inZZ`

Answer 2

`x=60°`
and
`x=300°`

Explanation:

`=>2cos x+4=5`

`=>2cos x=5-4`

`=>cos x=1/2`

`=> x=cos^(-1)(1/2)`

`=> x=60°` `(0°<=x<=90°)`

`=> x=300°` `(270°<=x<=360°)`