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mihailvagin199
@mihailvagin199
July 2022
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sin^2 2x/7 - 2 sin 2x/7 cos 2x/7 - 3 cos^2 2x/7=0 помогите решить , очень нужно (
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sedinalana
Разделим на cos²2x/7
tg²2x/7-2tg2x/7-3=0
tg2x/7=a
a²-2a-3=0
a1+a2=2 U a1*a2=-3
a1=-1⇒tg2x/7=-1⇒2x/7=-π/4+πn,n∈z⇒x=-7π/8+7πn/2,n∈z
a2=3⇒tg2x/7=3⇒2x/7=arctg3+πk,k∈z⇒x=3,5arctg3+3,5πk,k∈z
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Answers & Comments
tg²2x/7-2tg2x/7-3=0
tg2x/7=a
a²-2a-3=0
a1+a2=2 U a1*a2=-3
a1=-1⇒tg2x/7=-1⇒2x/7=-π/4+πn,n∈z⇒x=-7π/8+7πn/2,n∈z
a2=3⇒tg2x/7=3⇒2x/7=arctg3+πk,k∈z⇒x=3,5arctg3+3,5πk,k∈z