Sin^{2} x= /frac{1}{2} cos^{2} x= /frac{1}{9} tg^{2} x=1 ctg^{2}x=3 Решите пожалуйста подробно

1
sin²x=1/2
(1-cos2x)/2=1/2
1-cos2x=1
cos2x=0
2x=π/2+πn,n∈z
x=π/4+πn/2,n∈z
2
cos²x=1/9
(1+cos2x)/2=1/9
9+9cos2x=2
9cos2x=-7
cos2x=-7/9
2x=+-(π-arccos7/9)+2πn,n∈z
x=+-1/2(π-arccos7/9)+πn,n∈z
3
tg²x=1
tgx=-1⇒x=-π/4+πn,n∈z
tgx=1⇒x=π/4+πk,k∈z
4
ctg²x=3
ctgx=-√3⇒x=5π/6+πn,n∈z
ctgx=√3⇒x=π/6+πk,k∈z