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

##### 2 Answers

Mar 2, 2018

#### 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#

Mar 2, 2018

and