Question

4 и 5 задание. помогите срочно

Answer 1

1) - 4tq5x = 0;
-4≠0 ⇒tq5x =0;
5x =π*k 
x =π/5*k; k∈Z.
2) 2 -2ctq(2x+π/8) =0;
ctq(2x+π/8) =1;
2x+π/8 =π/4 +π*k ;
x =π/16 +π/2*k ; k∈Z.
3) tqx/2*(tqx -1) =0  ,  x∈( -3π/2;π/2) .
  [{tqx/2 =0; tqx  определен 
  [  tqx -1 =0; tqx/2  tопределен
tqx/2 = 0 ⇒x/2 =π*k⇔x₁ = 2π*k  k∈Z.
tqx =1⇒x₂ = π/4+π*k ,k∈ Z .
x = π*n ; n∈ Z.
ответ : - 3π/4 ; 0 ;  π/4 .
4)  sin(7x-π/3)/(1 -cos(7x-π/3)) =1/√3 ;
√3 sin(7x-π/3) = 1  -cos(7x-π/3;
cos(7x-π/3) +√3 sin(7x-π/3)    =1;
1/2*cos(7x-π/3) +√3/2*sin(7x-π/3) =1/2;
cosπ/3*cos(7x-π/3) +sinπ/3*sin(7x-π/3) =1/2;
cos(7x -2π/3) =1/2;
7x -2π/3 =(+/-)π/3 +2π*k ; k∈Z.
x =   (+/-)π/21 +2π/21 +2π/7*k  ; k∈Z.
x₁= - π/21+ 2π/21+2π/7*k ; k∈Z.
x₁=π/21+2π/7*k ; k∈Z.
x₂  = π/21+2π/21 +2π/7*k ; k∈Z.
x₂ =π/7+2π/7*k ; k∈Z.
π/21+2π/7*k ; π/7+2π/7*k ; k∈Z.
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5) arctqx+3/arctqx =4;
(arctqx)² -4arctqx  +3 =0;
arctqx  =1 ⇒ x=tq1;
arctqx  =3⇒ x=tq3.