Goniometrické rovnice
1. tg^2x + sin^2x + cos^2x = 2
2. (tgx + cotgx)^2 - (tgx - cotgx)^2 = 4
3. tgx = 2sinx
4. 4cos^2x = 3cotg^2x
5. 2sinx + tgx + 2cosx + 1 = 0
6. √2 cosx - cotgx - √2 sinx + 1 = 0
7. (1 + cos2x) sinx = 4cos^2x
Moc vás žádám o pomoc.
2. (tgx + cotgx)^2 - (tgx - cotgx)^2 = 4
3. tgx = 2sinx
4. 4cos^2x = 3cotg^2x
5. 2sinx + tgx + 2cosx + 1 = 0
6. √2 cosx - cotgx - √2 sinx + 1 = 0
7. (1 + cos2x) sinx = 4cos^2x
Moc vás žádám o pomoc.
Odpovědi
Diskuze
(jen abys pochopil, součet druhých mocnin sin a cos je rovno jedné)
tg^2(x) = 1
tg(x) = ±1
x = π/4 + kπ
x = (3/4)π + kπ
a jestli je y*y=1
pak je y=±1
a^2 - b^2 = (a-b)(a+b)
[tg(x) + cotg(x) - tg(x) + cotg(x)][tg(x) + cotg(x) + tg(x) - cotg(x)] = 4
2cotg(x) * 2tg(x) = 4 | :4
cotg(x) * tg(x) = 1
1/tg(x) * tg(x) = 1
1 = 1 => nekonečně mnoho řešení
Podmínka tg(x)≠0 => cosx≠0 => x≠ (2k+1)(π/2)
sin(x)/cos(x) = 2sin(x) |*cos(x)≠0 => x≠(2k+1)(π/2)
sin(x) = 2sin(x)cos(x)
2sin(x)cos(x) - sin(x) = 0
sin(x)[2cos(x)-1] = 0
1) sin(x) = 0 => x = kπ
2) 2cos(x) = 1 => cos(x) = 1/2 => x = π/3 + 2kπ
4cos^2(x) = 3cos^2(x)/sin^2(x) |* sin^2(x)≠0 => sin(x)≠0 => x≠kπ
4cos^2(x)sin^2(x) = 3cos^2(x)
4cos^2(x)sin^2(x) - 3cos^2(x) = 0
cos^2(x)[4sin^2(x)-3] = 0
1) cos^2(x) = 0 => cos(x) = 0 => x =(2k+1)(π/2)
2) 4sin^2(x)-3 = 0 => sin^2(x) = 3/4 => sin(x) = ±√3/2 => x=(π/3)+kπ nebo x=(2π/3)+kπ
√2cos(x) - √2sin(x) - cos(x)/sin(x) + 1 = 0 |*sin(x)≠0 => x≠kπ
√2cos(x)sin(x) - √2sin^2(x) - cos(x) + sin(x) = 0
√2 sin(x)[cos(x)-sin(x)] - [cos(x)-sin(x)] = 0
[cos(x)-sin(x)][√2 sin(x)-1] = 0
1) cos(x) - sin(x) = 0
cos(x) = sin(x) |* 1/cos(x) ≠0 => x≠(2k+1)(π/2)
1 = tg(x)
x = π/4 + kπ
2) √2sin(x) - 1 = 0 => sin(x) = 1/√2 => sin (x)= √2/2 => x=π/4+2kπ nebo x=3π/4+2kπ
2sin(x) + sin(x)/cos(x) + 2cos(x) + 1 = 0 |*cos(x)≠0 => x≠(2k+1)(π/2)
2sin(x)cos(x) + sin(x) + 2cos^2(x) + cos(x) = 0
sin(x)[2cos(x) + 1] + cos(x)[2cos(x) + 1] = 0
[2cos(x) + 1][sin(x)+cos(x)] = 0
1) sin(x) + cos(x) = 0
sin(x) = -cos(x) |* 1/cos(x)
tg(x) = -1 => x = 3π/4+kπ
2) 2cos(x) + 1 = 0
cos(x) = -1/2 => x = 2π/3+2kπ nebo x=4π/3+2kπ
[cos^2(x) + sin^2(x) + cos^2(x) - sin^2(x)] sinx = 4cos^2x
2cos^2(x)sin(x) = 4cos^2(x)
4cos^2(x) - 2cos^2(x)sin(x) = 0 | :2
2cos^2(x) - cos^2(x)sin(x) = 0
cos^2(x)[2 - sin(x)] = 0
1) cos^2(x) = 0 => cos(x) = 0 => x = (2k+1)(π/2)
2) 2 - sin(x) = 0
sin(x) = 2 nemá řešení