(3x² -19x+20)*(2cosx+√<span>3)=0 ;
--------------------------------------
Один из множителей равно нулю
a)
</span>3x² -19x+20 =0 ; D =19² -4*3*20 =361 -240 =121 =11²
x₁= (19 -11)/2*3 = 4/3 ;
x₂<span> =(19 +11) /6 =5.
b)
</span>2cosx+√<span>3=0
</span>cosx = - (√<span>3) /2 ;
</span>x = ±(π-π/6)+2πn ,n∈Z.
<span>
ответ : 4/3 , 5 ; </span> ±(<span>π-π/6)+2πn ,n∈Z.
</span><span>* * * * * * * * * * * * * * * * * * * * * * * * * *
Удачи Вам !
</span>
ОДЗ sinx≠0⇒x≠πn
2cos²x-cosx-1=0
cosx=a
2a²-a-1=0
D=1+8=9
a1=(1-3)/4=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn
a2=(1+3)/4=1⇒cosx=1⇒x=2πn,не удовл ОДЗ
х=-4π/3;-2π/3;2π/3;4π/3
18,9-6,8=12,1
12,1-5,2=6,9
6,9+4,1=11
А) х²+6х+8 = х²+4х+2х+8 = х(х+4)+2(х+4) = (х+4)(х+2)
б) х²-8х+15 = х²-3х-5х+15 = х(х-3)-5(х-3)= (х-3)(х-5)