1) 2) sinx-√3·cosx=√3 3)sinx·cosx=√2 - вопрос №3171136123332 от Не123098 22.03.2020 01:24

Question

Алгебра: 1) 2) sinx-√3·cosx=√3 3)sinx·cosx=√2

Не123098 · Answer

1)2tgx/(1-ctg²(π/2-x))=12tgx/(1-tg²x)=1tg2x=1⇒2x=π/4+πn⇒x=π/8+πn/22)2(1/2*sinx-√3/2*cosx)=√3sinx*cosπ/3-sinπ/3*cosx=√3/2sin(x-π/3)=√3/2x-π/3=(-1)^n *π/3+πnx=π/3+(-1)^n *π/3+πn3)sinxcosx=√21/2*sin2x=√2sin2x=2√2∉[-1;1]нет решения...