1)

2) sinx-√3·cosx=√3

3) sinx·cosx=√2

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