Question

Решите уравнение2^(sin2x/(cos2x-1))=0.5^√3

Answer 1

2^(sin2x/(cos2x-1))=2^(-sqrt(3))

sin2x/(cos2x-1)=-sqrt(3)

cos2x-1=cos^2x-sin^2x-sin^2x-cos^2x=-2sin^2x

2sinxcosx/-2sin^2x=-ctgx

-ctgx=-sqrt(3)

ctgx=sqrt(3)

x=П/6+Пk