6х = arctg(-3) + πn, n∈Z
6x=-arctg3 + πn, n∈Z
x= - (1/6)arctg3+(π/3)·n, n∈Z
О т в е т. - (1/6)arctg3+(π/3)·n, n∈Z
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ОТВЕТ: - 2 ; - 0,5
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X1+x2=-3a
x1*x2=a/2
x1²+x2²=(x1+x2)²-2x1*x2=(-3a)²-2*a/2=9a²-a=38
9a²-a-38=0
D=1+1368=1369
a1=(1-37)/18=-2
a2=(1+37)/18=19/9
3sin²x-7sinxcosx+4cos²x=0/cos²x
3tg²x-7tgx+4=0
tgx=a
3a²-7a+4=0
D=49-48=1
a1=(7-1)/6=1⇒tgx=1⇒x=π/4+πn,n∈z
a2=(7+1)/6=4/3⇒tgx=4/3⇒x=arctg4/3+πk,k∈z