Question 1102550
There must be an error or omission in the stated problem, because {{{ -1 <=  cos(x)   <= 1  }}}
cos(x) = 2  has no solution,  real or complex  x.  
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Here is what happens if you try to solve cos(z) = 2  for  z complex:
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{{{ cos(z) = (1/2)(e^(iz) + e^(-iz) ) }}}   <— definition 
Let z=a+bi  where a,b are real and i is the imaginary unit

{{{ cos(a+bi) = (1/2)(e^(i(a+bi)) + e^(-i(a+bi))) }}}
= {{{ (1/2)(e^(-b))(e^(ai)) + (1/2)(e^b)(e^(-ai)) }}}
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using {{{ e^(i*theta) = cos(theta) + i*sin(theta) }}}  <— Euler's equation
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= {{{ (1/2)(e^(-b))(cos(a) + i*sin(a)) + (1/2)(e^b)(cos(-a) + i*sin(-a)) }}}

Now setting this last equation to 2  you get (real part is 2, imaginary part is 0) .  Start by setting the imaginary part to 0:
Noting sin(-a) = -sin(a): 
{{{ (1/2)e^(-b)sin(a) - (1/2)(e^b)sin(a)  = 0 }}}
{{{    e^(-b) - e^b = 0 }}} ==>  b = 0

and for the real part:
{{{  (1/2)e^(-b)cos(a) + (1/2)(e^b)cos(-a) = 2 }}}
plugging in b=0, and noting cos(a) = cos(-a):
{{{   (1/2) cos(a) + (1/2)cos(a) = 2 }}}
 {{{   cos(a) = 2 }}}   which looks like we are back to the original problem….  EXCEPT now 'a' is REAL and we know this has no solution.
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