Решите:
(2/√3)(tgx-ctgx)=tg²x+ctg²x-2

(2/√3)(tgx-ctgx)=tg²x+ctg²x-2(2/√3)(tgx-ctgx)=(tgx-ctgx)²2/√3)(tgx-ctgx)-(tgx-ctgx)²=0(tgx-ctgx)(2/√3-tgx+ctgx)=0tgx-1/tgx=0(tg²x-1)/tgx=0(tgx-1)(tgx+1)=0,tgx≠0tgx=1⇒x=π/4+πk,k∈ztgx=-1⇒x=-π/4+πk,k∈z2/√3-tgx+1/tgx=02tgx-√3tg²x+√3=0tgx=a√3a²-2a-√3=0D=4+12=16a1=(2-4)/2√3=-1/√3⇒tgx=-1/√3⇒x=-π/6+πk,k∈za2=(2+4)/2√3=√3⇒tgx=√3⇒x=π/3+πk,k∈z