Задача 55809 1) 2 sinx cosx + 5 cos^2 x = 4 ...

5fbe40a564081d1dafcb1f06

25.11.2020 14:54:17

1) 2 sinx cosx + 5 cos^2 x = 4 2)sin2x - 3cos^3 x = 4

5f3ea7e3faf909182968ddd9

25.11.2020 15:05:32

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1)
1=sin^2x+cos^2x

4=4*(sin^2x+cos^2x)

2 sinx cosx + 5 cos^2 x = 4*(sin^2x+cos^2x)

2sinx*cosx+5cos^2x-4sin^2x-4cos^2x=0

2sinx*cosx+cos^2x-4sin^2x=0 - однородное тригонометрическое уравнение. Делим на cos^2x

4tg^2x-2tgx-1=0

D=4-4*4*(-1)=20

tgx=(2-2sqrt(5))/8 или tgx=(2+2sqrt(5))/8
tgx=(1-sqrt(5))/4 или tgx=(1+sqrt(5))/4
[b]x=arctg((1-sqrt(5))/4) +πk, k ∈ Z[/b] или [b]x=arctg((1+sqrt(5))/4) +πn, n ∈ Z[/b]

2) там точно [b]cos^3x[/b]