Solve the equation on the interval (0,2pi). Write your answer in exact simplest form.
1 answer:
9514 1404 393
Answer:
x ∈ {π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18}
Step-by-step explanation:
Using the trig identity ...
cos(2x) = 1 -2sin²(x)
we can rewrite the equation as ...
(1 -2sin²(3x)) -5sin(3x) +2 = 0
2sin²(3x) +5sin(3x) -3 = 0 . . . . . standard form
(sin(3x) +3)(2sin(3x) -1) = 0 . . . . factored form
There are no real values of x that will make the first factor zero. The second factor is zero when ...
sin(3x) = 1/2
3x = arcsin(1/2) +2nπ or (2n+1)π -arcsin(1/2)
x = 1/3(π/6 +2nπ) or 1/3((2n +1)π -π/6)
x = (12n+1)π/18 or (12n+5)π/18
x ∈ {π/18, 5π/18, 13π/18, 17π/18, 25π/18, 29π/18}
You might be interested in
This problem is very difficult to imagine without the figure. However I dug up on other sources and I think I found the correct figure to work with (see the attached pic).
We can see that in the figure, the hexagon is inscribing the pentagon. What is meant here to be lines of reflection simply means lines of symmetry. If we take on the hexagon alone, there are a lot of lines of symmetry. We can create the line intersecting E & B and the figure would still be symmetrical on both sides or D & A, F & C and etc. So there are a lot for hexagon alone.
However in this case, our lines of symmetry is made limited by the presence of the pentagon. If we slice the pentagon into two, the only line of symmetry we could create would be the line intersecting O and the median of LM. Other lines would not create a symmetrical half.
Therefore the line of reflection is only <u>1.</u>
