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??
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,
To find all solutions of the equation 4sin(x)cos(x) = 1 in the interval [0, 2π), you correctly started by simplifying the equation to 2sin(2x) = 1. Then you solved for 2x, which gave you the solutions π/6 and 5π/6.
However, by using the double angle formula, you can see that sin(2x) = 1/2 has additional solutions beyond the ones you found. The double angle formula states that sin(2x) = 2sin(x)cos(x). So, in order for sin(2x) = 1/2, either sin(x) = 1/2 and cos(x) = 1, or sin(x) = -1/2 and cos(x) = -1.
For sin(x) = 1/2, you can use the unit circle or trigonometric identities to find the solutions in the interval [0, 2π). The solutions for sin(x) = 1/2 are x = π/6 and x = 5π/6, which are the solutions you found initially.
However, for sin(x) = -1/2, you need to find the corresponding angle(s) in the interval [0, 2π). The solutions for sin(x) = -1/2 are x = 7π/6 and x = 11π/6.
Since you were looking for solutions in the interval [0, 2π), you correctly did not include x = 7π/6 and x = 11π/6 initially. However, the answer key you referred to included these solutions because it considered the larger interval [0, 4π) as the solution set.
Therefore, if you're considering the interval [0, 4π), then the full set of solutions for 4sin(x)cos(x) = 1 is x = π/6, 5π/6, 7π/6, and 11π/6. But if you're specifically looking in the interval [0, 2π), then the solutions are x = π/6 and 5π/6 only.
It's important to consider the given interval when determining the full set of solutions for an equation involving trigonometric functions.