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Как графический установить сколько решений имеет интервал (-\pi ; 2\pi )

a. sin(x) = \frac{1}{2}
b. cos(x) = -\frac{1}{3}
c. 2sin(x) = \sqrt{2}

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Ответ:
Mtzv17
Mtzv17
09.07.2021

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4,8(19 оценок)
Ответ:
znmbuzinis
znmbuzinis
09.07.2021

АСДК - трапеция, основания АС=12 см и ДК=4 см

АВ = 12-4 = 8 см

АК = 12+4 = 16 см

По Пифагору

ВК² = АК²-АВ² = 16²-8² = 256-64 = 3*64

ВК = 8√3 см

∠ВАК = arccos(АВ/АК) = arccos(1/2) = 60°

∠ВКА = 90 - ∠ВАК = 30°

∠ДКА = ∠ВКА + 90 = 120°

Полная площадь трапеции

S(ACDK) = 1/2(AC+DK)*BK = 1/2(12+4)*8√3 = 64√3 см²

Площадь сектора большого круга (серая штриховка)

S₁₂ = πR²/360*α = π*12²*60/360 = π*12*12/6 = 24π  см²

Площадь сектора малого круга (зелёная штриховка)

S₄ = πR²/360*α = π*4²*120/360 = π*16/3 = 16π/3  см²

И площадь странной фигуры около касательной

S =  S(ACDK) - S₁₂ - S₄ =  64√3 -  24π -  16π/3  см²

S =  64√3 -  88π/3  см²


15 . две окружности, имеющие радиусы 4 и 12 см, внешне касаются, ab-их общая касательная. найдите пл
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