-2x²-5x-x+3>0|*(-1)
2x²-6x+3<0
D=(-6)²-4*2*3=36-24=12=√12=3,464
x1=(-(-6)+3,464)/2*2=9,464/4
x1=2,366
x2=(-(-6)-3,464)/4=2,536/4
x2=0,634
x∈(0,634; 2,366)
Sin 9° cos 99° - sin99° cos9°<span> =
</span>= sin 9° cos (90°+9°) - sin(90° +9°) cos9° =
= sin 9° (- sin 9°) - cos9° cos9° =
= - sin² 9° - cos² 9° =
= - (sin² 9° +
cos² 9°) = - 1
Приравниваешь к 0
y=0
1,5x-3=0
1,5x=3
x=2
При решение 1-4 используем в основном формулы приведения
1.
a) cos(-210°)=cos210°=cos(180°+30°)=-cos30°=-√3/2
б) tg(4π/3)=tg(3π/2-π/6)=ctgπ/6 =√3
2.
a) sin(3π/2-α)-cos(π+α)=-cosα+cosα=0
б) tg(π+α)-ctg(π/2-α) =tgα-tgα=0
в) sin2α+(sinα-cosα)^2= 2sinαcosα+sin^2α-2sinαcosα+cos^2α= 1
3.
a) sin(π-2α)/(1+cos2α)=tgα
sin2α/(sin^2α+cos^2α+cos^2α-sin^2α)=tgα
2sinαcosα/2cos^2α=tgα
sinα/cosα=tgα
tgα=tgα
б) 4sinαcosα/(cos^2α-sin^2α) = 2tg2α
2*2sinαcosα/cos2α=2tg2α
2sin2α/cos2α=2tg2α
2tg2α=2tg2α
4.
a) (ctgα-tgα)tg2α (1)
Учитывая, ctgα=1/tgα и tg2α=2tgα/(1-tg^2α)
Подставим в (1)
(ctgα-tgα)tg2α = (1/tgα-tgα)*2tgα/(1-tg^2α)=2tgα(1-tg^2α)/tgα(1-tg^2α)=2
б)(1+cos2α)/sin(π/2-α)=(sin^2α+cos^2α+cos^2α-sin^2α)/cosα = =2cos^2α/cosα=2cosα
5.
sin(290°+α)-cos(340°-α)/sin(110°+α)=-2
sin(270°+(20°+α))-cos(360°-(20°+α))/sin(90°+(20°+α))=-2
-cos((20°+α)-cos((20°+α)/cos((20°+α) =-2
-2cos((20°+α)/cos((20°+α)=-2
-2 = -2