Odpowiedź:
[tex]\huge\boxed{x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\ \vee\ x=\dfrac{\pi}{12}+k\pi,\ k\in\mathbb{C}}[/tex]
Szczegółowe wyjaśnienie:
[tex]\cos^2\left(2x-\dfrac{\pi}{6}\right)-\cos\left(\dfrac{\pi}{6}-2x\right)=0\\\\\cos^2\left(2x-\dfrac{\pi}{6}\right)-\cos\bigg[-\left(2x-\dfrac{\pi}{6}\right)\bigg]=0\\\\\cos^2\left(2x-\dfrac{\pi}{6}\right)-\cos\left(2x-\dfrac{\pi}{6}\right)=0\\\\\cos\left(2x-\dfrac{\pi}{6}\right)\left(\cos\left(2x-\dfrac{\pi}{6}\right)-1\right)=0\\\Updownarrow\\\cos\left(2x-\dfrac{\pi}{6}\right)=0\ \vee\ \cos\left(2x-\dfrac{\pi}{6}\right)-1=0[/tex]
[tex]\cos\left(2x-\dfrac{\pi}{6}\right)=0\Rightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\qquad|+\dfrac{\pi}{6}\\\\2x=\dfrac{3\pi}{6}+\dfrac{\pi}{6}+k\pi\\\\2x=\dfrac{4\pi}{6}+k\pi\\\\2x=\dfrac{2\pi}{3}+k\pi\qquad|:2\\\\\boxed{x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}}\ k\in\mathbb{C}[/tex]
[tex]\cos\left(2x-\dfrac{\pi}{6}\right)-1=0\qquad|-1\\\\\cos\left(2x-\dfrac{\pi}{6}\right)=1\Rightarrow2x-\dfrac{\pi}{6}=2k\pi\qquad|+\dfrac{\pi}{6}\\\\2x=\dfrac{\pi}{6}+2k\pi\qquad|:2\\\\\boxed{x=\dfrac{\pi}{12}+k\pi}[/tex]
