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