inequality: solve sin(x - pi/2) + 1 < 0.5 for 0 < x &l

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Math wiz ya rite 09

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Hello! I already mentioned this in a previous topic, but anyways, I bought an SAT practice book for getting ready for the big test. It was pretty well formated, except it only has the answers/explanations to random questions. I was just hoping that you guys could help me with a question here or there that I didn't know how to solve. Thanks!

Find the set of all real x, where 0 < x < 2pi, such that sin(x - pi/2) + 1 < 0.5.
 
To start, subtract the 1 from both sides. Then find where sin(ß) = -1/2 for angles ß between, say, -pi/2 and 2pi. This will divide the interval [-pi/2, 2pi] into subintervals.

Look at the sine curve, and determine on which subintervals the sine curve is below (less than) y = -1/2. This solves the inequality for ß. Then plug "x - pi/2" in for ß, and add pi/2 to the ends of the subintervals.

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.
 
Remember the unit circle?

sin(x - pi/2) + 1 < 0.5

sin(x-pi/2)<-0.5

Where does sin(x) equal -1/2? 7pi/6 and 11pi/6

7pi/6<x-pi/2<11pi/6

5pi/3<x<7pi/2
 
Math wiz ya rite 09 said:
i got 0 < x < pi/2 i just graphed this to get my answer. Seemed the simplest thing to do for me? am i right??
Click to expand...
As the worked solution (provided by the tutor in a previous reply) showed, your answer is not correct. So try following the step-by-step instructions (provided by a still-earlier reply) to work out the answer instead.

If you cannot obtain the correct solution, please reply showing all of your work and reasoning. Thank you.

Eliz.
 
Exactly...so, within my range of values, what range is between 0 and 2pi. That is, what values between 5pi/3 and 7pi/2 do fall between 0 and 2pi?
 
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