1) 2) sinx-√3·cosx=√3 3)sinx·cosx=√2

Question

1)


2) sinx-√3·cosx=√3


3)sinx·cosx=√2

Answer 1

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