1
√cos2x*(tg2x-1)=0
ОДЗ
cos2x≥0
x∈[-π/4+πk/2;π/4+πk/2,k∈z]
[cos2x=0⇒x=-π/4+πk/2 U x=π/4+πk/2
[tg2x=-π⇒2x=π/4+πk⇒x=π/8+πk/2
ответ x={-π/4+πk/2;π/8+πk/2;π/4+πk/2,k∈z}
2
cosx/sinx-sin2x=0
cosx(1-2sin²x)/sinx=0
sinx≠0⇒x≠πk,k∈z
cosx=0⇒x=π/2+πk
-π≤π/2+πk≤π/2
-2≤1+2k≤1
-3≤2k≤0
-3/2≤k≤0
k=-1 x=π/2-π=-π/2
k=0 x=π/2
1-2sin²x=0
sin²x=1/2
sinx=-1/√2⇒x=-π/4+2πk U x=-3π/4+2πk
sinx=1/√2⇒x=π/4+2πk U x=3π/4+2πk
-π≤-π/4+2πk≤π/2
-4≤-1+8k≤2
-3≤8k≤3
-3/8≤k≤3/8
k=0 x=-π/4
-π≤-3π/4+2πk≤π/2
-4≤-3+8k≤2
-1≤8k≤5
-1/8≤k≤5/8
k=0 x=-3π/4
-π≤π/4+2πk≤π/2
-4≤1+8k≤2
-5≤8k≤1
-5/8≤k≤1/8
k=0 x=π/4
-π≤3π/4+2πk≤π/2
-4≤3+8k≤2
-7≤8k≤-1
-7/8≤k≤-1/8
нет решения