Question

cu explicaţie va rog :)

Answer 1

cos²x+sin²x=1, cos²x=1-sin²x
inlocuim in ecuatia data
1-sin²x+sinx=5/4
-sin²x+sinx-1/4=0
sin x=t
-t²+t-1/4=0
t²-2x1/2sint+(1/2)²=0
(t-1/2)²=0
t=1/2, sinx=1/2
x=(-1)ⁿarcsin(1/2)+πn, n∈Z
n=0, x=(-1)⁰(π/6)+0=π/6
n=1, x=(-1)¹(π/6)+π=5π/6
n=2, x=(-1)²(π/6)+2π=π/6+2π∉[0,2π]
n=-1, x=(-1)⁻¹(π/6)-π=-π/6-π∉[0,2π]