f(x) = $\frac{1}{2}(2cosx- cos2x)- \frac{\sqrt{3}}{2} (sin 2x + 2sinx)$ , tìm max min của f(x) với x

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$f(x)=(cosx-\sqrt{3}sinx)-(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x)$

$=-2sin(x-\dfrac{\pi}{6})-cos(2x-\dfrac{\pi}{3})$

$=-2sin(x-\dfrac{\pi}{6})-1+2sin^2(x-\dfrac{\pi}{6})$

$=...$ (PT bậc 2 ---> tìm đc max, min)
 
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