sin²x+3sinx-4=0
t=sinx
t²+3t-4=0
D=9+16=25 √D=5
t₁=(-3+5)/2=1
t₂=(-3-5)/2=4
sin x=1
x=(π/2)+2πn, n∈Z
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sin²x+3sinx-4=0
t=sinx
t²+3t-4=0
D=9+16=25 √D=5
t₁=(-3+5)/2=1
t₂=(-3-5)/2=4
sin x=1
x=(π/2)+2πn, n∈Z