Задача 33613 4^(cos^2(x+Pi/4)) = 2*2^(cosx), [4Pi;...

slava191

2019-02-16 15:49:31

4^(cos^2(x+Pi/4)) = 2*2^(cosx), [4Pi; 11Pi/2]

sova

2019-02-16 16:09:37

★

2^(2cos^2(x+(π/4)))=2^(1+cosx)
2cos^2(x+(π/4))=1+cosx;
По формуле: 2cos^2α=1+cos2α
1+cos(2x+(π/2))=1+cosx
cos(2x+(π/2))=cosx

По формулам приведения
cos( α +(π/2))=-sin α

-sin2x=cosx

2sinx*cosx+cosx=0
cosx*(2sinx+1)=0

cosx=0 или 2sinx+1=0
x=(π/2)+πn, n ∈ Z или sinx=-1/2; x=(-1)^(k)*(-π/6)+πk, k ∈ Z

Отрезку [4π; 11π/2] принадлежат корни:
(π/2)+4π=9π/2;
(π/2)+5π=11π/2
(-5π/6)+6π=31π/6

О т в е т.
а)(π/2)+πn, n ∈ Z ; (-1)^(k)*(-π/6)+πk, k ∈ Z
б)9π/2;11π/2;31π/6