2cos^2(x) - 3cos(x) = -1.


The way to solve such equations is to introduce new variable  y = cos(x).

Then your equation takes the form


    2y^2- 3y = -1,   or


    2y^2 - 3y + 1 = 0.


Apply the quadratic formula


     =  =  = .


So, there are two roots


    1)   y =  =  = 1.


         Recall that y = cos(x).  So,  cos(x) = 1.

         It implies  x = ,  k = 0, +/-1, +/-2, . . . 



    2)   y =  =  = .


         Recall that y = cos(x).  So,  cos(x) = 1/2.

         It implies  x = ,  k = 0, +/-1, +/-2, . . . 

              or     x = ,  k = 0, +/-1, +/-2, . . . 


ANSWER.  x = ,  k = 0, +/-1, +/-2, . . .     or

         x = ,  k = 0, +/-1, +/-2, . . .  or

         x = ,  k = 0, +/-1, +/-2, . . .