$ (2cos x +1)^2 = 2cos x + 4cos^2 x $
$ 4 cos^2 x + 4 cos x + 1 = 2cos x + 4cos^2 x $
$ 2cos x + 1 = 0 $
$ cos x = -\frac{1}{2} $
$ x = \pm \frac {2\pi}{3} + 2k\pi; \qquad k \in \mathbb{Z}$
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