Find all solutions to the given equation in the interval [0,2π). Give the exact solution, including "pi" for π. For any unused answer boxes, enter DNE in all capital letters.

Question

(a) 2cos x=2
so cos x=1 =0..(now what?)



(b) 4cos x+2=0
cos x = -1/2 (120)
cos is negative in quads 2 and 3

not sure what Iḿ doing..thank you

help please...I donẗ get this..

Answer 1

2cosx = 2
cosx = 1
you should familiarize yourself with the basic sine and cosine curves,
looking at the cosine curve notice that when x = 0 , cos0 = 1
and when x = 2π, cos (2π) = 1
so your solution is x = 0, 2π for your required interval.

if all else fails, get your calculator, set it to Rad and enter
2ndF, cos, 1 , =
and you will get 0, so x = 0 is a solution
you know the period of the cosine curve is 2π, so 0+2π will be your
next solution.

(b) 4cos x+2=0
cos x = -1/2, you are right to say x must be in quads II and III
you should also know that cos 60° = +1/2 , but we have -1/2
so in quad II, your x must be 180-60 or 120°
in quad III, your x must be 180+60 or 240°
in radians that would be 2π/3 and 4π/3

so x = 2π/3, 4π/3