Вопрос пользователя:
1)2cos в квадрате x + 5cos x=3;
2)tg x – 3ctg x=0;
3)sin 3x-sin x=0;
4)2 sin x + sin 2 x=0
Если не трудно,буду благодарен:)
Илюха отвечает:
1) заменим cosx = a
2a² + 5a – 3 =0
-5±√25+24 /4 = a1 = -3, а2 = 1/2
а1 не подходит.
cosx = 1/2
x = ±π/3 + 2πn
2) tgx – 3ctgx = 0, tgx – 3/tgx = 0, tg²x-3 = 0, 1-cos2x/1+cos2x = 3
4cos2x = -2, сos2x = -1/2
2x = ±π-arccos1/2 +2πn
2x = ± π-π/3 + 2πт = 2π/3 + 2πт
x = ±π/3 + πn
3) Sin3x – Sinx = 0
2cos(3x+x/2)*sin(3x-x/2)
2cos2xsinx = 0
cos2xsinx = 0
1) cos2x = 0 2) sinx = 0
2x = π/2+πn x = πn
x = π4+πn/2
4) 2sinx +sin2x = 0, 2sinx +2sinxcosx = 0 , 2sinx(1+cosx) = 0
1) 2sinx = 0 2) 1+cosx = 0
sinx = 0 cosx = -1
x = πn x = π + 2πn