1
sinx*cosx=0,5sin2x
√0,5sin2x(1/tg2x+1)=0
ОДЗ
{sin2x≥0⇒2πn≤2x≤π+2πn⇒πn≤x≤π/2+πn
{tg2x≠0⇒2x≠πn⇒x≠πn/2
x∈(πn;π/2+πn),n∈z
sin2x=0⇒x=πn⇒x=πn/2,n∈z∉ОДЗ
1/tg2x+1=0⇒1/tg2x=-1⇒tg2x=-1⇒2x=-π/2+πn⇒x=-π/4+πn/2,n∈z
4
sin(3π/2+x)=-cosx
2cos²x-√3cosx=0
cosx(2cosx-√3)=0
cosx=0⇒x=π/2+πn,n∈z
-7π/2≤π/2+πn≤-π
-7≤1+2n≤-4
-8≤2n≤-5
-4≤n≤2,5
n=-4πx=π/2-4π=-7π/2
n=-3⇒x=π/2-3π=-5π/2
2cosx-√3=0⇒2cosx=√3⇒cosx=√3/2⇒x=-π/6+2πn,n∈z U x=π/6+2πn,n∈z
-7π/2≤-π/6+2πn≤-2π
-21≤-1+12n≤-12
-20≤12n≤-11
-5/3≤n≤-11/12
n=-1⇒x=-π/6-2π=-13π/6
-7π/2≤π/6+2πn≤-2π
-21≤1+12n≤-12
-22≤12n≤-13
-11/6≤n≤-13/12
нет решения
Ответ x={-7π/2;-5π/2;-13π/6}