А2.
а)
cos(x/2) =-1/2 ;
x/2 =± 2π/3 +2πn ;
x = ± π/3 +πn ; n∈N.
---
б)
sin(x+π/4) =1/2 ;
[x+π/4 =π/3 +2πn ; x+π/4 =π-π/3 +2πn , n∈N.
[x =π/12 +2πn ; x=5π/12 +2πn , n∈N.
-------
А3.
а)
tq²x -3tqx -4 =0 ; * * * tqx =-1* * *
[ tqx =-1 ; tqx =4 ⇔ [x = -π/4+ πn ,x =arctq4+πn, n∈N.
---
б)
cos2x +cos²x +sinxcosx =0 ;
(cosx-sinx)(cosx+sinx) +cosx(cosx+sinx) =0 ;
(cosx+sinx)(cosx-sinx+cosx) =0 ; || cosx ≠0 иначе ⇒и sinx =0 ||
(tqx+1)(2 -tqx) =0 ;
[tqx = -1 ; tqx =2.⇒[x = -π/4+πn; x =arctq2+πn, n∈N.
или -------
cos2x +cos²x +sinxcosx =0 ;
2cos²x+sinxcosx -sin²x =0 ; || :(-cos²x)
tq²x -tqx -2 =0 ⇒[tqx = -1 ; tqx =2.⇒[x = -π/4+πn; x =arctq2+πn, n∈N.
---
в)
4sin²x -cosx -1 =0 ;
4(1-cos²x) -cosx -1 =0 ;
4cos²x +cosx -3 =0 ;
[cosx = -1 ; cosx =3/4 ⇒[ x =(2n+1)π; x =±acrcos(3/4)+2πn ,n∈N.
-------
А4.
sinx +cosx =√2 ;
sinx*1/√2+cosx*1/√2 =1 ;
sinx*cos(π/4)+cosx*sin(π/4)=1 ;
sin(x+π/4) =1 ;
x+π/4 =π/2 +2πn ,n∈Z⇔ x =π/4 +2πn ,n∈Z.
---------------
В1.
sin2x+3 =3sinx +3cosx ⇔sin²x+cos²x +2sinxcosx +2=3(sinx+cosx) ;
(sinx+cosx)² -3(sinx+cosx)+2 =0 ⇔ t² -3t +2 =0; ||t=sinx+cosx=t||.
[ t =2 ;t=1⇔ [sinx+cosx =2(не имеет решения); sinx+cosx =1.
sinx+cosx =1⇔√2*sin(x+π/4) =1 ⇔sin(x+π/4)=1/√2⇒
[x+π/4=π/4+2πn , x+π/4=π-π/4+2πn ,n∈Z.
⇔[x=2πn , x=π/2+2πn ,n∈Z.
---------------
В2.
cos4x*sin5x -cos5x*sin4x =1;
sin(5x-4x) =1 ;
sinx=1⇒x =π/2+2πn ,n∈Z.
------
УДАЧИ !