Question

Sa se rezolve in R ecuatiile:

Answer 1

recomandat cand avem sinx si cos x in ac.ecuatie; impartim toata relatia cu cos(x/3)...rezulta tgx/3=-1
x/3=-pi/4+kπ
de unde x=-3π/4+3kπ,  k apartine Z ;
nu e o contradictie tgx/3 are perioada principala (-3π/2;3π/2)

b) impartim cu  4cosx
 rezulta tgx=3/4
x= kπ+arctg3/4


c)  e o ecuatieomogemna mascata...impartim cu cos^2 (x0
 rezulta tg^2 (x)-3-2tgx=0
 t²-2t-3=0
t1,2=(2+-√16)/2
t1=-1
t2=3

tgx=-1
x= -π/4+kπ

tgx=3
x=arctg3+kπ

x∈(-π/4+kπ)∪(arctg3+kπ)


d)
4sin²x+2sinxcox-3=0
sin²x+3sin²x-3+2sinxcox=0
sin²x+3(sin²x-1)+2sinxcox=0

sin²x-3cos²x+2sinxcosx=0
ecuatie omogena impartim cu cos²x si notam tgx=5
t²+2t-3=0
t1=1=tgx ⇒x=kπ+π/4
t2=-3=tgx⇒x= kπ+arctg(-3)= kπ-arctg3
x∈(kπ+π/4)∪ (kπ-arctg3),k∈Z