Đáp án:
cos2x=1/2$$\\$$ =>cos2x=cos\frac{pi}{3}$$\\$$ =>[(2x=pi/3 + k2pi),(),(2x=-pi/3 + k2pi):} (k in ZZ)$$\\$$ =>[(x=pi/6 + kpi),(),(x=-pi/6 + kpi):} (k in ZZ)$$\\$$ Vậy...$$\\$$
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Công thức: cosu=cosv$$\\$$ =>[(u=v + k2pi),(),(u=-v+ k2pi):} (k in ZZ)$$\\$$

cos 2x = \frac{1}{2}
<=> cos 2x = cos \frac{\pi}{3}
<=> 2x = \pm \frac{\pi}{3} + k.2\pi (k in ZZ)
<=> x = \pm \frac{\pi}{6} + k.2\pi (k in ZZ)
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