Post a New Question

trig

posted by on .

Find the exact solutions of the equation in the interval [0,2pi).

sin(x/2)+cos(x)=0

  • trig - ,

    recall the half-angel conversions
    cos 2A = 1 - 2sin^2 A
    so
    cos x = = 1 - 2sin^2 (x/2)

    sin(x/2) + 1 - 2sin^2 (x/2) = 0
    2sin^2(x/2) - sin(x/2) - 1 = 0
    (2sin(x/2) + 1)(sin(x/2) - 1) = 0
    sin(x/2) = -1/2 or sin(x/2) = 1
    x/2 = 7π/6 or x/2 = 11π/6 or x/2 = π/2
    x = 7π/3 or x = 11π/3 or x = π (420° , 660· , 180° )

    so for the given domain
    x = π

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question