Find all the solutions and indicate solutions in interval [0,2π)?

sin2x - sinx cosx = cosx

i need to see the work, that would help a lot.

3 Answers

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  • 8 years ago
    Favorite Answer

    is the first sin term of 2x or it is square of sin x?

  • 8 years ago

    Ok Greg--

    Using the double angle identity for sin(2x) you'll get 2sin(x)cos(x)

    Now subtract and bring over the cos(x)and set = 0 and you'll get

    cos(x)*(sin(x) -1) = 0.

    By ZPP cos(x) = 0 when x = pi/2 and 3pi/2

    and sin(x) = 1 when x = pi/2

    Source(s): Math tutor for several years.
  • Kate
    Lv 4
    8 years ago

    sin 2x - sin x cos x = cos x

    Substitute in sin2x = 2sinxcosx [from sin(a + b) = sin a cos b + cos a sin b]:

    2sin x cos x - sin x cos x = cos x

    sin x cos x - cos x = 0

    Factor out cos x:

    cos x(sin x - 1) = 0

    cos x = 0

    x = π/2, 3π/2

    sin x = 1

    x = π/2

    ∴ x = π/2, 3π/2

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