Ответы на вопрос:
(2/√3) (tgx-ctgx) = tg2x+ctg2x-2
(2/√3) (tgx-ctgx) = (tgx-ctgx) 2
2/√3) (tgx-ctgx) - (tgx-ctgx) 2=0
(tgx-ctgx) (2/√3-tgx+ctgx) = 0
tgx-1/tgx=0
(tg2x-1) / tgx=0
(tgx-1) (tgx+1) = 0, tgx≠0
tgx=1⇒x=π/4+πk, k∈z
tgx=-1⇒x=-π/4+πk, k∈z
2/√3-tgx+1/tgx=0
2tgx-√3tg2x+√3=0
tgx=a
√3a2-2a-√3=0
D=4+12=16
a1 = (2-4) / 2√3=-1/√3⇒tgx=-1/√3⇒x=-π/6+πk, k∈z
a2 = (2+4) / 2√3=√3⇒tgx=√3⇒x=π/3+πk, k∈z
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