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

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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