cos4x+2cos^{2} x=0
Va rog am nev urgenta!
Folositi formule din trigonometrie de cls a 10 a daca se poate :)


Răspuns :

cos4x=cos²2x-sin²2x=2cos² 2x-1
expresia devine
2cos² 2x-1+2cos²2x=0

4cos²2x=1

cos²2x=1/4

cos2x=1/2
2x= 2kπ+/-arccos(1/2)=2kπ+/-π/3
x=kπ+/-π/6

cos2x=-1/2
2x=2kπ+/-arccos(1/2)=2kπ+/- 2π/3
x=kπ+/-π/3


deci x∈(kπ+/-π/6)∪(kπ+/-π/3), k∈Z