Trig

Find all solutions in the interval [0,2pi)

4sin(x)cos(x)=1

2(2sinxcosx)=1

2sin2x=1

2x=1/2

x= pi/6, and 5pi/6
Then since its 2x i divided these answers by 2 and got pi/12 and 5pi/12

However, when i checked the answer key there solutions 13pi/12 and 17pi/12 were included suggesting the interval [0,4pi). Why are these solutions included??

asked by Martha
  1. ok up to
    2sin2x=1
    then:
    sin 2x = 1/2
    so 2x = π/6 or 2x = 5π/6 <--- you had that. good!

    x = π/12 or x = 5π/12 <-- again, you had that, now....

    the period of sin 2x = 2π/2 = π
    so adding π to any existing answer will yield a new answer,
    x=π/12+π = 13π/12
    x = 5π/12 + π = 13π/12
    and there are your other two answers,

    posted by Reiny

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