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Math Help - Finding the domain

  1. #1
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    Finding the domain

    The question:
    Find the domain of:
    \sqrt{1 - 2sinx}

    My attempt:
    1 - 2sinx = 0
    -2sinx = -1
    2sinx = 1
    sinx = \frac{1}{2}

    I worked out that x = 30 degrees, or Pi/6

    So I want the values where sinx is larger than or equal to 1/2:
    \frac{\pi}{6} + 2\pi k <= x <= \frac{5\pi}{6} + 2\pi k
    Where k is an integer.

    I thought this made sense, but the answer in my book is:
    -\frac{7\pi}{6} + 2\pi k <= x <= \frac{\pi}{6} + 2\pi k

    What am I doing wrong? :/
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  2. #2
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    Hello, Glitch!

    The book's answer is wrong . . .


    Find the domain of: . f(x) \:=\:\sqrt{1 - 2\sin x}

    Use inequalities . . .

    . . 1 - 2\sin x \:\ge \:0 \quad\Rightarrow\quad \text{-}2\sin x \:\ge \:\text{-}1 \quad\Rightarrow\quad \sin x \:\le \frac{1}{2}


    If \sin x is between \text{-}\frac{1}{2} and \frac{1}{2}

    . . then x is basically between \text{-}\frac{\pi}{6} and \frac{\pi}{6}


    Therefore: . \text{-}\frac{\pi}{6} + 2\pi k \;\leq \;x \;\leq\; \frac{\pi}{6} + 2\pi k

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  3. #3
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    How did you get the -1/2?

    Also, was my answer incorrect?
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  4. #4
    Member Mathelogician's Avatar
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    Hi,
    Glitch, Your solution is where the sine is greater than or equal to 1/2! Infact you should find the angles which their sine is less than or equal to 1/2.
    The simplest way is to use a unit circle or a sine graph to solve this Trigonometric inequality.
    1-Draw a unit circle.
    2-Draw a horizontal line to cosine axis such that intersects the sine axis at the point 1/2.
    3-Since at the 1st and 2nd quadrants sine is 1/2 at pi/6 and pi-pi/6=5pi/6, the line intersects the circle at those points. Now you can see that all the angles between pi/6 and 5pi/6 have sines greater than 1/2. So what interval do you think you should takeas solution?
    Therefore you can see that the book's solution is true! (Note that -7pi/6 is in the same location of 5pi/6 and so they have the equal Trig relationships)
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  5. #5
    Member Mathelogician's Avatar
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    And Soroban, you should note that when sine<=1/2 then it could has all the values in [-1,1/2]. Infact you forgot the numbers in [-1,-1/2).
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  6. #6
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    Thank you very much! As you say, I was looking at when it was larger than 1/2 like a fool. I got the answer the textbook states.

    My method was to draw the sine graph and get an idea of what was happening. Hopefully that's a satisfactory method.
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