решить уравнения
1)(2/√3) (tgx-ctgx) = tg2x+ctg2x-2

(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