<span>а) sin α =√3/2α=(-1)ⁿ+πn, n∈Z
б) cos α = - √2/2α=+-3π/4+2πn, n∈Z
в) tg α = √ 3α=π/3+πn, n∈Z
г) ctg α = -1α=3π/4+πn, n∈Z
вычислите:а) tg²α + ctg²α=tg²α + ctg²α+2ctgα*tgα-2ctgα*tgα=(tgα+ctgα)²-2ctgα*tgα=(tgα+ctgα)²-2=3²-2=7
б)(3*sin α - 4*cos α)/(5*sin α + 6*cos α)
tgα=-3sinα/cosα=-3sinα=-3cosα(3*sin α - 4*cos α)/(5*sin α + 6*cos α)=(3(-3cosα)-4cosα)/(5(-3cosα)+6cosα) =-13cosα/(-9cosα)=13/9
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arcsin √2/2 - arcos0 + (arctg √ 3)/ (arcctg√3/ 3)=π/4-π/2+π/3:π/3=1-π/4
<span>х^4-4*х^2- 5 =(x^2-5)*(x^2+1)=(x-5^0,5)*(x+5^0,5)*(x^2+1)=0; x1=5^0,5; x2=-5^0,5;</span>
<span>3х+у=7
9х-4у=-7
1)</span><span>3х+у=7
y=7-3x
Подставляем во второе:
2)</span><span>9х-4у=-7</span>
9х-4(7-3х)=-7
9х-28+12х=-7
21х=21
х=1
у=7-3х=7-3=4
Ответ:х=1;у=4.