Question

cosx?tgx=2 и 0<X<пи2

Answer 1

$1+tg^2x=frac{1}{cos^2x}$

$cos^2x=frac{1}{1+tg^2x}$

$cosx=sqrt{frac{1}{1+tg^2x}}$

$cosx=sqrt{frac{1}{1+2^2}}=sqrt{frac{1}{1+4}}=sqrt{frac{1}{5}}=frac{sqrt{5}}{5}$