x∈IR:cos⁡2x+2cos2⁡x−2=0=

cos⁡2x+2cos2⁡x−2=0=cos2⁡(π/6)⇒x=nπ±π/6

x∈IR:cos⁡2x+2cos2⁡x−2=0=

cos⁡2x+2cos2⁡x−2=0=cos2⁡(π/6)⇒x=nπ±π/6

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$\{x \in I R : \cos 2 x + 2 \cos^{2} x - 2 = 0\} =$

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$\{n π \pm \frac{π}{6} : n \in Z\}$

$\{2 n π - \frac{π}{3} : n \in Z\}$

$\{n π \pm \frac{π}{3} : n \in Z\}$

$\{2 n π + \frac{π}{3} : n \in Z\}$

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$\cos 2 x + 2 \cos^{2} x - 2 = 0$

$= \cos^{2} (π / 6) \Rightarrow x = n π \pm π / 6$

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x∈IR:cos⁡2x+2cos2⁡x−2=0=

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$\{x \in I R : \cos 2 x + 2 \cos^{2} x - 2 = 0\} =$