1. 2tg^2x+4 cos^2x=7 2. 4sin^2x-2sinx=√3(-1+2sinx)
Moc prosím o pomoc.
Odpovědi
Diskuze
(2(1-cos^2(x)))/cos^2(x) + 4cos^2(x) = 7
s: cos^2(x) = a
(2 - 2a)/a + 4a = 7 |*a≠0
2-2a + 4a^2 = 7a
4a^2 - 9a + 2 = 0
a_1,2 = (9±√(81-32))/8 = (9±7)/8
a_1 = 2 => cos^2(x) = 2 => ∅
a_2 = 1/4 => cos^2(x) = 1/4 => cos(x) = ±1/2 => x_1 = π/3 + 2kπ, x_2 = 2π/3 + 2kπ, x_3 = 4π/3 + 2kπ, x_4 = 5π/3 + 2kπ
Podmínky: cos(x)≠0 => x≠(2k+1)*π/2
2sin(x)*(2sin(x)-1) - √3(2sin(x)-1) = 0
(2sin(x)-1)(2sin(x)-√3) = 0
1) 2sin(x)-1 = 0
sin(x) = 1/2 => x_1 = π/6 + 2kπ, x_2 = 5π/6 + 2kπ
2) 2sin(x) - √3 = 0
2sin(x) = √3
sin(x) = √3/2 => x_3 = π/3 + 2kπ, x_4 = 2π/3 + 2kπ