Question

Pentru ce valoare a lui x

$\dfrac{cos \ x}{tg \ 30}+sin\ x$

are valoare maxima?

(30 este in grade)

Answer 1

$P=\frac{cosx}{tg30^o}+sinx \\\frac{dP}{dx}=0<=>(\frac{cosx}{tg30^o}+sinx )'=0\\(\frac{cosx}{tg30^o}+sinx )'=\frac{(cosx)'tg30^o-(tg30^o)'cosx}{tg^2x}+cosx=\frac{-sinx*tg30^o-0*cosx}{tg^230^o}+cosx= \frac{-\frac{\sqrt{3}}{3}sinx }{(\frac{\sqrt{3}}{3})^2 }+cosx =\frac{-\frac{\sqrt{3}}{3}sinx }{\frac{1}{3})}+cosx=cosx-\sqrt{3}sinx$

$\frac{dP}{dx}=0=>cosx-\sqrt{3}sinx=0<=>cosx=\sqrt{3}sinx|:sinx=>\frac{cosx}{sinx}=\sqrt{3}=>ctgx=\sqrt{3}$

$ctgb=a,a\in R=>b=arcctga+\pi n,n\in Z,arcctga\in(0,\pi)$

$ctgx=\sqrt{3}=>x=arcctg{\sqrt3}+\pi n,n\in Z=>x=\frac{\pi}{6}+\pi n,n\in Z$

$Pt.n=0=>x=\frac{\pi}{6}+0= \frac{\pi}{6}=>x=30^o$