1
ОДЗ
tgx<0<br>x∈(-π/2+2πk;π+2πk,k∈z) U (3π/2+2πk;2π+2πk,k∈z)
sin2x-2sinx-cosx+1=0
2sinx(cosx-1)-(cosx-1)=0
(cosx-1)(2sinx-1)=0
cosx-1=0⇒cosx=1⇒x=2πk,k∈z+ОДЗ нет решения
2sinx-1=0⇒sinx=1/2⇒x=π/6+2πk U x=5π/6+2πk,k∈z +ОДЗ⇒
x=5π/6+2πk,k∈z
2
ОДЗ
sin(x-π/4)>0⇒2πk1-2sin²x+sinx=0
sinx=t
2t²-t-1=0
D=1+8=9
t1=(1-3)/4=-1/2⇒sinx=-1/2⇒x=7π/6+2πk,k∈z U x=11π/6+2πk,k∈z+ОДЗ ⇒
х=7π/6+2πk,k∈z
t2=(1+3)/4=1⇒sinx=1⇒x=π/2+2πk ∈ОДЗ
1)0≤7π/6+2πk≤2π
0≤7+12k≤12
-7≤12k≤5
-7/12≤k≤5/12
k=0⇒x=7π/6
2)0≤π/2+2πk≤2π
0≤1+4k≤4
-1≤4k≤3
-1/4≤k≤3/4
k=0⇒x=π/2
7π/6+π/2=(7π+3π)/6=10π/6=5π/3
Ответ 5π/3