2sin²x+5sinxcosx+5cos²x=1;...

Тема в разделе "Алгебра", создана пользователем westernborn, 10 янв 2010.

  1. westernborn

    westernborn New Member

    2sin²x+5sinxcosx+5cos²x=1;
     
  2. dr.Twister

    dr.Twister New Member

    (1)
    2 cos²x = 1 + sin x
    2 - 2 sin²x = 1 + sin x
    2 sin²2 + sin x - 1 = 0
    D = 1 + 8 = 3²
    sin x = (1 + 3) / 4 = 1
    sin x = (1 - 3) / 4 = - 1/2

    (2)
    2 sin²x - 5 sin x cos x + 5 cos²x = 1
    2 sin²x - 5 sin x cos x + 5 cos²x = sin²x + cos²x
    sin²x - 5 sin x cos x + 4 cos²x = 0 | /cos²x
    tg²x - 5 tg x + 4 = 0
    (tg x - 4) (tg x - 1) = 0
    tg x = 4, tg x = 1

    (3)
    sin x + cos x + sin 3x = 0
    sin x + cos x + sin (2x + x) = 0
    sin x + cos x + sin 2x cos x + sin x cos 2x = 0
    sin x (1 + cos 2x) + cos x (1 + sin 2x) = 0
    sin x (1 + cos²x - sin²x) + cos x (1 + sin 2x) = 0
    sin x (2 cos²x) + cos x (1 + sin 2x) = 0
    cos x (2 sin x cos x + 1 + sin 2x) = 0
    cos x (2 sin 2x + 1) = 0
    cos x = 0
    x = ± pi / 2 + 2 pi k, k in Z
    2 sin 2x + 1 = 0
    sin 2x = - 1/2
    2x = - pi / 6 + (-1)^m 2 pi m, m in Z
    x = - pi / 12 + (-1)^m pi m

    (5)
    √3 cos x - sin x = 1 | /2
    (√3/2) cos x - (1/2) sin x = 1/2
    cos (pi/6) cos x - sin (pi/6) sin x = 1/2
    cos (pi/6 + x) = cos (pi/3)
    pi/6 + x = ± pi/3 + 2 pi k, k in Z
    x = ± pi/3 - pi/6 + 2 pi k
    (4)
    √3 cos x - sin x = 0
    √3 cos x = sin x
    tg x = √3
    x = pi / 3 + pi k, k in Z
     

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