УМОЛЯЮ, ПОМОГИТЕ ПОЖАЛУЙСТА!!!! 4 примера решить
1) COS 2x=√3
2) Sin 2x= - √3/2
3) tg (x-π/6)=1/√3
4) Sin x/2 = -√2
1) COS 2x=√3
2) Sin 2x= - √3/2
3) tg (x-π/6)=1/√3
4) Sin x/2 = -√2
More Questions From This User See All
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
2)sin2x=-√3/2;⇒
2x=(-1)^k·(-π/3)+kπ;k∈Z; x=(-1)^k·(-π/6)+kπ;k∈Z;
3)tg(x-π/6)=1/√3=√3/3;⇒
(x-π/6)=π/6+kπ;k∈Z; x=π/6+π/6+kπ=π/3+kπ;k∈Z;
4)sinx/2=-√2;⇒√2≈1.41>1;<-1sinα<1;решения нет