findall exact solutions ( in terms of radians) to the equation 2cos(40)-radical 3 =0

Answer 1

There is no variable in your equation, so there is nothing to solve for.

You can rewrite it as
cos(40) = (sqrt3)/2

but that is not true, whether the 40 is in degrees or radians.

Answer 2

is that 40 meant to say 4(theta) ?

If so, I will use 4x instead of 4theta for easier typing

2cos 4x - √3 = 0
cos 4x = √3/2
from my 30-60-90 triangle, I know cos 30 = √3/2
and the cosine is positive in quadrants I and IV

so 4x = 30 degrees or 4x = 330 degrees
x = 7.5 degrees or x = 82.5 degrees

in radians
x = pi/24 or 11pi/24

the period of cos 4x is pi/2 or 90 degrees
so by adding multiples 90 or pi/2 to each answer we can generate more answers

x = 7.5, 97.5, 187.5, 277.5, ...
or x = 82.5, 172.5, 262.5, ....

general solution:
x = pi/24 + k(pi/2) or 11pi/4 + k(pi/2), where k is an integer.