Question 1151815
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<pre>

    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


    {{{y[1,2]}}} = {{{(3 +- sqrt(3^2 - 4*2*1))/(2*2)}}} = {{{(3 +- sqrt(1))/4}}} = {{{(3 +- 1)/4}}}.


So, there are two roots


    1)   y = {{{(3 + 1)/4}}} = {{{4/4}}} = 1.


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

         It implies  x = {{{2k*pi}}},  k = 0, +/-1, +/-2, . . . 



    2)   y = {{{(3 - 1)/4}}} = {{{2/4}}} = {{{1/2}}}.


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

         It implies  x = {{{pi/3 + 2k*pi}}},  k = 0, +/-1, +/-2, . . . 

              or     x = {{{5pi/3 + 2k*pi)}}},  k = 0, +/-1, +/-2, . . . 


<U>ANSWER</U>.  x = {{{2k*pi}}},  k = 0, +/-1, +/-2, . . .     or

         x = {{{pi/3 + 2k*pi}}},  k = 0, +/-1, +/-2, . . .  or

         x = {{{5pi/3 + 2k*pi)}}},  k = 0, +/-1, +/-2, . . . 
</pre>

Solved.


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