Question

rozwiąż równanie: sinx · cos2x = 0, x∈<-pi; pi>

Answer 1

sinx=0 v cos2x=0

x=-π v x=0 v x=π

cos2x=0

2x=π/2+2kπ v 2x=-π/2+2kπ /:2, k∈C

x=π/4+kπ v x=-π/4+kπ

Odp. x∈{-π, -3π/4, -π/4,0, π/4. 3π/4, π}