Question

How do you find all the angles between 0 and 2 pi which satisfy 3cosx-2=0?

Answer 1

`0.2677pi and 1.732pi`, nearly

Explanation:

`cos x =2/3>0`. x is in the 1st and 4th quadrants.

In the 1st quadrant, #x = cos^(-1)(2/3)=48.19^o= (48.19/180)pi radian =

0.841 radian= `(0.842/3.14159)pi=0.2677pi` nearly

#cos (2pi-x)=cos x = 2/3

So, x in the fourth quadrant `= (2pi-0.841)=5.442=(5.442/3.1459)pi=1.732pi` radian