Даю 35 баллов. Помогите решить номера А2-А4

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) 4sin ²x -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 !