Find all solutions of the equation in the interval [0, 2pi). 2 cos 0 - 13 = 0
Write your answer in radians in terms of t.
If there is more than one solution, separate them with commas.

The solutions of 2·cos(θ) - √3 = 0in the interval [0, 2pi) are;

π/6,  13/6·π

Step-by-step explanation:

The given that the equation is 2·cos(θ) - √3 = 0

The solution of the above equation in the interval [0, 2pi) are required

Therefore, the domain includes 0 ≤ θ < 2pi

2·cos(θ) - √3 = 0

2·cos(θ)  = √3

cos(θ)  = √3/2

Therefore;

θ = cos⁻¹(√3/2)

The values are;

Where the domain is 0 ≤ θ < 2pi, we have;

π/6,  13/6·π

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)