Practice!

Sample Problem

Find all solutions of 2-sin2x=2sin2(x/2) in the interval [0, 2π). Check all that apply.

x=0

x=π/2

x=π

x=3π/2

Solution

2-sin2x=2sin2(x/2)

2-sin2x=2

2-sin2x=

2-sin2x=1-cosx

2-(1-cos2x)=1-cosx

1+cos2x-1+cosθ=0

cos2x+cosx=0

cos x(cosx+1)=0

cos x=0 OR cosx+1

cos x=0 : x=π/2, 3π/2

cos x+1=0: cosx=-1 à x=π

x=π/2, π, 3π/2