Question 1183076
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Find the solutions of the equation cos 2x + cos 𝑥 + 1 = 0 in the interval [0, 2𝜋).
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Your starting equation is

    cos(2x) + cos 𝑥 + 1 = 0    (1)


Use  cos(2x) = 2cos^2(x) - 1.   Substitute it into the given equation.  You will get


    (2cos^2(x) - 1) + cos(x) + 1 = 0,   or

     2cos^2(x) + cos(x) = 0.


Factor left side


    (2cos(x) + 1)*cos(x) = 0.


So,  EITHER  cos(x) = 0, giving  x = {{{pi/2}}},  {{{3pi/2}}},


     OR      2cos(x) + 1 = 0,  giving  cos(x) = - {{{1/2}}},   x = {{{2pi/3}}},  {{{4pi/3}}}.


<U>ANSWER</U>.  The solutions are  {{{pi/2}}},  {{{3pi/2}}},  {{{2pi/3}}},  {{{4pi/3}}}.
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Solved.