1.
a) cosx = -1/2 ;
x = ±2π/3+2πn ,n∈Z.
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b) tq2x =(√3)/3;
2x =π/6+πn ,n∈Z;
x =π/12+(π/2)*n ,n∈Z.
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c) ctq(x -π/3) =√3 ;
x -π/3 =π/6+πn ,n∈Z;
x =π/2+πn ,n∈Z.
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3.
a)√3(tqx)² -tqx =0 ;
3(tqx)* (tqx -1/√3) =0 ;
[ tqx=0 ; tqx =1/√3 .⇒ [ x =πn ; x =π/6 +πn , n∈Z.
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b) 2(cosx)² - 5cosx +2 =0 ;
* * * замена переменной t =cosx ; -1 ≤ t ≤ 1 * * *
2t² - 5t +2 =0 ;
D =5² -4*2*2 =9 =3²
t₁=(5-3)/(2*2) =1/2⇒cosx =1/2 ⇔x = ±π/3+2πn ,n∈Z.
t₂=(5+3)/4 =2 не корень исходного уравнения
c) 3sinx +4cosx =0 ;
sinx = -(4/3)cosx ; \cosx≠0 ;
tqx = -4/3 ;
x = - arctq(4/3) +πn , n∈Z.