anonymous
  • anonymous
Studying for Precalculus Test, and teaching myself is not working... The question states, "Find all solutions in the interval 0 degrees
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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johnweldon1993
  • johnweldon1993
\(\large 1 - 2sin\theta = 0\) on interval \(\large 0 \le \theta \le 2\pi\) Focus first on the equation \[\large 1 - 2sin\theta = 0\] Subtract 1 from both sides and divide everything by -2 \[\large sin(\theta) = \frac{1}{2}\] So now I will ask you...when does \(\large sin(\theta) = \frac{1}{2}\) ?
anonymous
  • anonymous
Oh! Would it be \[\pi/6, 5\pi/6, 7\pi/6, and 11\pi/6?\]
johnweldon1993
  • johnweldon1993
Yes and No..Remember we only want it within the boundaries of 0 and 2pi Which of those are within this boundary :)

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anonymous
  • anonymous
Uhm... well 2pi is the whole circle right? So... I'm confused lol
johnweldon1993
  • johnweldon1993
Lol okay lets try and do this out Imagine going completely around the circle JUST ONCE!!! because going around once means you have gone 2pi The 2 places where sin(theta) = 1/2 are at pi/6 and 5pi/6 as shown |dw:1447189252139:dw| There are no more parts where sin(theta) = 1/2 unless we start to go around the circle again
anonymous
  • anonymous
Ohhh okay! So when a question states interval of 0 to 2pi, its only the parts on half of the circle?
johnweldon1993
  • johnweldon1993
Not quite, but I didnt point that out So our unit circle is just |dw:1447189601533:dw|
johnweldon1993
  • johnweldon1993
Note here...we start where I labeled 2pi and move counter clockwise around the circle until we get back to that same point |dw:1447189670067:dw| That distance we just traveled going once fully around the circle...is 2pi
anonymous
  • anonymous
Ohh, and the other points would be negative, so you wouldnt include them... right?
johnweldon1993
  • johnweldon1993
So where they say the interval is 0 to 2pi....we start at that 2pi label....and go fully around the circle once back to that point...that is our interval For example If the interval were instead 0 to pi/2 ....that would only be from |dw:1447189820254:dw| in that little area of the circle
johnweldon1993
  • johnweldon1993
And correct...since sin is only positive where I have shaded: |dw:1447189899414:dw|
anonymous
  • anonymous
Fantastic ! Thank you so very much!
johnweldon1993
  • johnweldon1993
No problem :)

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