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Math Help - Trigonemetric Functions - Need help understanding constraints

  1. #1
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    Unhappy Trigonemetric Functions - Need help understanding constraints

    Hello,

    Can someone please help me understand how the answers were obtained?

    1. (sqrt3)tanx+1=0 -----> tanx = -1/(sqrt3) or 30 deg.

    answer is: x = 5pi/6 + kpi

    2. 4(cos^2)x-1=0 -----> cosx = 1/2 or 60 deg.

    answer is: x = pi/3 + kpi, 2pi/3 + kpi

    3. 3(csc^2)x-4=0 -----> cscx=2/(sqrt3) or 60 deg.

    answer is: x = pi/3 + kpi, 2pi/3 + kpi

    .................

    I can do all the algebraic steps, what I'm confused about is how did they come up with kpi as the answer shouldn't it be 2kpi?

    because this is what i got...

    1. [5pi/6 + 2kpi, 11pi/6 + 2kpi]
    what I was thinking was that since tanx=-1/(sqrt3) I would need the solutions of the quadrant where tan is negative, which is quadrant II and IV

    2. [pi/3 + 2kpi, 5pi/3 + 2kpi]
    again, since cosx=1/2, which is positive...the solutions would be in quadrant I and IV

    3...same idea, positive answer cscx = 2/(sqrt3)..solutions are in quadrant I and II

    Please help, thanks!
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  2. #2
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    Can someone please help me understand how the answers were obtained?

    1. (sqrt3)tanx+1=0 -----> tanx = -1/(sqrt3) or 30 deg.

    answer is: x = 5pi/6 + kpi

    2. 4(cos^2)x-1=0 -----> cosx = 1/2 or 60 deg.

    answer is: x = pi/3 + kpi, 2pi/3 + kpi

    3. 3(csc^2)x-4=0 -----> cscx=2/(sqrt3) or 60 deg.

    answer is: x = pi/3 + kpi, 2pi/3 + kpi

    .................

    I can do all the algebraic steps, what I'm confused about is how did they come up with kpi as the answer shouldn't it be 2kpi?

    because this is what i got...

    1. [5pi/6 + 2kpi, 11pi/6 + 2kpi]
    what I was thinking was that since tanx=-1/(sqrt3) I would need the solutions of the quadrant where tan is negative, which is quadrant II and IV

    2. [pi/3 + 2kpi, 5pi/3 + 2kpi]
    again, since cosx=1/2, which is positive...the solutions would be in quadrant I and IV

    3...same idea, positive answer cscx = 2/(sqrt3)..solutions are in quadrant I and II

    Please help, thanks!
    1. The period of the tangent function is \pi, not 2\pi.


    So you have

    \sqrt{3}\tan{x} + 1 = 0

    \sqrt{3}\tan{x} = -1

    \tan{x} = -\frac{1}{\sqrt{3}}


    Now since your solution is in the second and fourth quadrants:

    x = \left\{\pi - \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\} + k\pi, k \in \mathbf{Z}

    x = \left\{ \frac{5\pi}{6}, \frac{11\pi}{6}\right\} + k\pi, k \in \mathbf{Z}.


    2. 4\cos^2{x} - 1 = 0

    4\cos^2{x} = 1

    \cos^2{x} = \frac{1}{4}

    \cos{x} = \pm \frac{1}{2}


    Since the solution can be in any of the four quadrants:


    x = \left\{\frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}

     = \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}.



    3. 3\csc^2{x} - 4 = 0

    3\csc^2{x} = 4

    \csc^2{x} = \frac{4}{3}

    \sin^2{x} = \frac{3}{4}

    \sin{x} = \pm \frac{\sqrt{3}}{2}.


    Since the solution can be in any four of the quadrants:


    x = \left\{ \frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}

     = \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\} + 2k\pi, k \in \mathbf{Z}.
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