Conditii de existenta a radicalilor: x≥ 1
Aplicam in stanga si in dreapta functia tangenta si avem:
tg(arccos √x/2)= tg(arctg √x-1)
sin(arccos √x/2)/ cos( arcos √x/2)= √x-1
√1-x/4/ √x/2= √x-1
(√4-x)/√x= √x-1
ridicam la patrat si obtinem:
(4-x)/x= x-1
4-x= x²-x
x²=4
x=2
(x=-2 nu corespunde deoarece x≥1)
Proba:
arccos √2/2= arctg √1
pi/4= pi/4
Doamne ajuta!
