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1vaaann
@1vaaann
July 2022
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Помогите решить алгебру
Sin2x-2корня из трех •sin^(2) x+4 cosx-4корня из трех •sinx=0
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kalbim
Verified answer
Sin(2x) = 2*sinx*cosx
2*sinx*cosx - 2*sin^2(x) * sqrt3 + 4cosx - 4*sinx*sqrt3 = 0
(2*sinx*cosx + 4cosx) - 2sinx*sqrt3 *(sinx + 2) = 0
2cosx*(sinx + 2) - 2sinx*sqrt3 *(sinx + 2) = 0
(sinx + 2)*(2cosx - 2sinx*sqrt3) = 0
1) sinx + 2 = 0 - нет решений, т.к. синус не равен -2
2) 2cosx - 2sinx*sqrt3 = 0
2sinx*sqrt3 = 2cosx
tgx = (sqrt3)/3
x = arctg((sqrt3)/3) + pi*k
x = pi/6 + pi*k
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Answers & Comments
Verified answer
Sin(2x) = 2*sinx*cosx2*sinx*cosx - 2*sin^2(x) * sqrt3 + 4cosx - 4*sinx*sqrt3 = 0
(2*sinx*cosx + 4cosx) - 2sinx*sqrt3 *(sinx + 2) = 0
2cosx*(sinx + 2) - 2sinx*sqrt3 *(sinx + 2) = 0
(sinx + 2)*(2cosx - 2sinx*sqrt3) = 0
1) sinx + 2 = 0 - нет решений, т.к. синус не равен -2
2) 2cosx - 2sinx*sqrt3 = 0
2sinx*sqrt3 = 2cosx
tgx = (sqrt3)/3
x = arctg((sqrt3)/3) + pi*k
x = pi/6 + pi*k