Question

a) (36^sinx)^cosx = 6^sqrt2 sinx

b) корни на отрезке [2pi ; 7pi/2]

Answer 1

а) $(36^{sinx})^{cosx}=6^{sqrt2sinx} \ 6^{2sinxcosx}=6^{sqrt2sinx} \ 2sinxcosx=sqrt2sinx \ 2sinxcosx-sqrt2sinx=0 \ sinx(2cosx-sqrt2)=0$

$sinx=0 \ x=pi k, k in Z$ или $cosx=frac{sqrt2}{2} \ x=бfrac{pi}{4}+2pi n, n in Z$

 

б) На [2П; 7П/2]:

1) $2pileq pi k leq frac{7pi}{2} \ 2leq k leq 3,5 \ k=2; 3; x_1=2pi, x_2=3pi.$

2) $2pileq бfrac{pi}{4}+2pi n leq frac{7pi}{2} \ 2leq бfrac{1}{4}+2n leq 3,5$

$2,25leq 2n leq 3,75 \ 1,125leq n leq 1,875$ или $1,75leq 2n leq 3,25 \ 0,875leq n leq 1,625 \ n=1 \ x_3=frac{9pi}{4}$

Ответ: 2П; 3П; 9П/4.