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Math Help - Solving Linear Trig Equations?

  1. #1
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    Solving Linear Trig Equations?

    Hello again!

    I need some help on solving linear trig equations!


    1. 1 - tan (2x + pi/2) = 0 0 Θ ≤ 2 pi

    I was told that there are 4 answers possible for this question. How is this possible?

    2. Find all solutions of each equation.
    a) cot x = 2.3 (I can find it with special angles, but not this)
    b) 5 cos 3x = -3

    3. c) 2sinx - 1 = -2

    Thanks!
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  2. #2
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    Quote Originally Posted by finalfantasy View Post
    Hello again!

    I need some help on solving linear trig equations!


    1. 1 - tan (2x + pi/2) = 0 0 Θ ≤ 2 pi
    So,
    \tan (2x+\pi/2) = - 1.
    But, \tan (2x+\pi/2) = - \cot x.
    That means,
    -\cot 2x = - 1.
    Thus,
    \cot 2x = 1.
    I find it easier to think in tangents so take reciprocal of both sides,
    \tan 2x = 1
    Thus,
     2x = \frac{\pi}{4}+\pi n for interger n.
    Thus,
    x = \frac{\pi}{8} + \frac{\pi}{2}n = \frac{5\pi}{8}n.
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  3. #3
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    1) 1-tanx(2x + pi/2) = 0

    -tan(2x + pi/2) = -1

    tan(2x + pi/2) = 1

    tan is positive in Quad 1 and 3, using the CAST rule.

    Quadrant 1: tan(alpha)=1
    (alpha) = pi/4 (using special triangles)
    therefore (2x + pi/2) = pi/4
    2x = pi/4 - pi/2
    2x = -pi/4
    x = -pi/8 + pi(n)

    Quadrant 3: 2x + pi/2 = pi + pi/4 (to get to Quad 3 we must add pi to alpha)
    2x = pi + pi/4 - pi/2
    2x = 3pi/4
    x = 3pi/8 + pi(n)

    These answers are not between 0 and 2pi, therefore:

    We can sub 0, 1, and 2 for n for both quadrants.

    n= 0 : x = 3pi/8 (-pi/8 is NOT in domain)

    n= 1 : x = -pi/8 + pi
    = 7pi/8

    x = 3pi/8 + pi
    x = 11pi/8

    n= 2 : x = -pi/8 + 2pi
    x = 15pi/8

    Those are your four solutions. Hope that helped.
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  4. #4
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    Quadrant 3: 2x + pi/2 = pi + pi/4 (to get to Quad 3 we must add pi to alpha)
    2x = pi + pi/4 - pi/2
    2x = 3pi/4
    x = 3pi/8 + pi(n)


    Thank you all!

    It makes sense now, but how is 3pi/8 not between 0 and 2 pi? Does not it equal 67.5 deg?
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  5. #5
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    Quote Originally Posted by finalfantasy View Post
    Quadrant 3: 2x + pi/2 = pi + pi/4 (to get to Quad 3 we must add pi to alpha)
    2x = pi + pi/4 - pi/2
    2x = 3pi/4
    x = 3pi/8 + pi(n)


    Thank you all!

    It makes sense now, but how is 3pi/8 not between 0 and 2 pi? Does not it equal 67.5 deg?
    You're right, it actually is. I just worded what I meant wrong. I meant to say that not both solutions are between 0 and 2pi. That's why i did use 3pi/8 when subbing in n. Im glad that helped though! Its the unit we're doing right now as well.
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  6. #6
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    ^ when do you know you should sub n in for 0, 1 or 2 if the angle is in between 0 and 2 pi?

    If one out of the 2 answers from that angle is in between that interval, you would sub in n for that? as seen when you added 3pi /8? I'm terribly slow at getting this, but what does that do? Give you the ... coterminal, related angle ? :S


    And can you help me with this one too ^.^

    2. Find all solutions of each equation.
    a) cot x = 2.3 (I can find it with special angles, but not this)
    b) 5 cos 3x = -3

    3. c) 2sinx - 1 = -2

    You people here are very kind! Thanks!
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  7. #7
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    Quote Originally Posted by finalfantasy View Post
    ^ when do you know you should sub n in for 0, 1 or 2 if the angle is in between 0 and 2 pi?

    If one out of the 2 answers from that angle is in between that interval, you would sub in n for that? as seen when you added 3pi /8? I'm terribly slow at getting this, but what does that do? Give you the ... coterminal, related angle ? :S


    And can you help me with this one too ^.^

    2. Find all solutions of each equation.
    a) cot x = 2.3 (I can find it with special angles, but not this)
    b) 5 cos 3x = -3

    3. c) 2sinx - 1 = -2

    You people here are very kind! Thanks!
    Well, basically, when the question just asks to solve, then your domain is not restricted and you can do it for all numbers, thus putting the " + pi(n) " in there. Like I did here:

    x = -pi/8 + pi(n)
    x = 3pi/8 + pi(n)

    That means that it could go on for infinity, since the sine or cos curve keeps going, it never stops. UNLESS, there was a domain restriction. Then that tells you that you must only find the solutions on the curve between, say, 0 and 2pi. To do so, you sub in 0 for n giving you 0pi, 1 for n giving you pi, and 2 for in giving you 2pi. Those are the values between 0 and 2pi. And that's how you get your solutions. Hope that helps a little bit.

    And for those other questions, I would love to help you, but seeing as I really dont have time right now, I can't. But an alternative to the special triangles is just doing a quick sketch of the curve. That helps you see where the points are exactly. I have to go study for my math test tomorrow, hope that helps!
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