Sin^2 (x) = 1/2

cos^2 (x) = 1/9

tg^2 (x) = 1

ctg^2 (x) = 3

Помогите решить

Sin^2 (x) = 1/2

1) Sin (x) = - 1/√2 2) Sin (x) = 1/√2 ⇒x=π/4+πn/2 n∈Z

cos^2 (x) = 1/9

1) cos (x) = - 1/3 2) cos (x) = 1/3 ⇒

1) x=arccos (-1/3) + 2πn = π-arccos (1/3) + 2πn=-arccos (1/3) + (2n+1) π

x = - arccos (-1/3) + 2πn=π+arccos (1/3) + 2πn=arccos (1/3) + (2n+1) π

2) (x) = arccos (1/3) + 2πn x) = - arccos (1/3) + 2πn

n∈Z

tg^2 (x) = 1

1) tg (x) = - 1 2) tg (x) = 1⇔x=π/4+πn/2 n∈Z

ctg^2 (x) = 3

1) ctg (x) = - √3 ⇔ tg (x) = - 1/√3 x=-π/6 + πn n∈Z

⇔

2) ctg (x) = √3 ⇔tg (x) = 1/√3 x=π/6 + πn n∈Z