Question 1155030
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<pre>

It is equivalent to THIS equation


    {{{(2sin(x)-1)^2}}} = 0


which, in turn, is equivalent to


    sin(x) = {{{1/2}}},


having the roots


    x = {{{pi/6}}} + {{{2k*pi}}},  k = 0, +/-1, +/-2, . . . 

and/or

    x = {{{5pi/6}}} + {{{2k*pi}}},  k = 0, +/-1, +/-2, . . . 
</pre>

Solved.