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naumova29
@naumova29
July 2022
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sedinalana
Verified answer
5
sin2x=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒sin2x=1/2⇒2x=π/6+2πk U x=5π/6+2πk⇒x=π/12+πk U x=5π/12+πk,k∈z
a2=(3+1)/4=1⇒sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z
6
y`=(6√(t²+1) -6t*2t/2√(t²+1))/(t²+1)=(6*(t²+1)-6t²)/√(t²+1)³=(6t²+6-6t²)/√(t²+1)³=
=6/√(t²+1)³
t=√3
6/√(3+1)³=6/8=3/4=0,75
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Answers & Comments
Verified answer
5sin2x=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒sin2x=1/2⇒2x=π/6+2πk U x=5π/6+2πk⇒x=π/12+πk U x=5π/12+πk,k∈z
a2=(3+1)/4=1⇒sin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z
6
y`=(6√(t²+1) -6t*2t/2√(t²+1))/(t²+1)=(6*(t²+1)-6t²)/√(t²+1)³=(6t²+6-6t²)/√(t²+1)³=
=6/√(t²+1)³
t=√3
6/√(3+1)³=6/8=3/4=0,75