Пусть 27^tg^2x=t>0 ⇒27^(-tg^2x)=1/t⇒t+81/t=30⇒
t^2-30t+81=0⇒по теореме Виетта
t1+t2=30; t1*t2=81⇒
t1=3; t2=27
1) t=3⇒27^(tgx)^2=3⇒3^(3(tgx)^2)=3⇒3(tgx)^2=1⇒
(tgx)^2=1/3⇒tgx=+(-)1√3⇒
x=π/6+πn; x=-π/6+πn
2) 1) t=27⇒27^(tgx)^2=27⇒(tgx)^2=1⇒tgx=+(-)1⇒
x=π/4+πn; x=-π/4+πn
Ответ:
x=π/6+πn; x=-π/6+πn
x=π/4+πn; x=-π/4+πn