Tìm GTLN, GTNN `y=cos(2x+2pi/3)+cos2x`

$y=cos(2x+\dfrac{2π}{3})+cos2x$

  $=cos2xcos(\dfrac{2π}{3})-sin2xsin(\dfrac{2π}{3})+cos2x$

  $=-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x+cos2x$

  $=\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x$

  $=\sqrt{{{(\dfrac{1}{2})}^{2}+{(\dfrac{\sqrt{3} }{2})}^{2}}}sin(2x+arcos\frac{\dfrac{-\sqrt{3}}{2}}{\sqrt{{{(\dfrac{1}{2})}^{2}+{(\dfrac{\sqrt{3} }{2})}^{2}}}})$

  $=sin(2x+\dfrac{5π}{6})$

$=>ymax=1$; $ymin=-1$

$ymax$ $tại$ $2x+\dfrac{5π}{6}=\dfrac{\pi}{2}+2k\pi=>x=-\dfrac{\pi}{6}+k\pi$

$ymin$ $tại$ \(\left[ \begin{array}{l}2x+\dfrac{5π}{6}=-\dfrac{1}{2}+2k\pi\\2x+\dfrac{5π}{6}=\dfrac{3}{2}+2k\pi\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=-\dfrac{2\pi}{3}+k\pi\\x=\dfrac{\pi}{3}+k\pi\end{array} \right.\)

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