SOLUTION: 1. Solve: 3 𝑐os^2𝑥 − 5 cos𝑥 − 2 = 0 i. 0 ≤ 𝑥 < 2𝜋 ii Over all real numbers. 2. Solve: sin 2𝑥 − 2 cos²𝑥 = 0 over the set of real numbers.

(1)  3*𝑐os^2(𝑥) − 5*cos𝑥 − 2 = 0.     (1)

     is a quadratic equation relative cos(x).  Use the quadratic formula to find cos(x)

         cos(x) =  =  = .


     The two roots are  cos(x) =  = 2  and  cos(x) =  =  = .


     cos(x) = 2 does not have solution, so we can forget about it.

     cos(x) =  in the interval 0 <= x <   has two solutions

              x =  = 1.9106332  and  x =  -  = 4.3725521 (approximate values).


     Over the set of all real numbers,  equation (1) has two families of solutions

              x =  = 1.9106332 +   and  x =  -  =  - 4.372552064,

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



(2)  sin 2𝑥 − 2 cos²𝑥 = 0.   (2)

     2*sin(x)*cos(x) - 2*cos^2(x) = 0,

     2*cos(x)*(sin(x) - cos(x)) = 0.


     It deploys in two equations.  First equation is

        cos(x) = 0.

     Over the set of all real numbers, it has solutions  x = ,  k = 0, +/-1, +/-2, . . . 


     Second equation is

         sin(x) - cos(x) = 0,  which is equivalent to

         sin(x) = cos(x),  or  tan(x) = 1.


         It has the solutions  x = ,  k = 0, +/-1, +/-2, . . . 


      ANSWER.  Over all real numbers, equation (2) has solutions  x =   and/or  x = ,  k = 0, +/-1, +/-2, . . .