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Math Help - Solving Trig Question

  1. #1
    Super Member
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    Solving Trig Question

    Hi
    I am having trouble solving the following equation:
    tan(3x-\frac{\pi}{6})
    range 0,2\pi
    this is what a have done, someone tell me where my mistake is.
    3x-\frac{\pi}{6}=\pi-\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi-\frac{\pi}{6},4\pi-\frac{\pi}{6},5\pi-\frac{\pi}{6},6\pi-\frac{\pi}{6}
    3x=\frac{5\pi}{6}+\frac{\pi}{6},\frac{11\pi}{6}+\f  rac{\pi}{6},\frac{17\pi}{6}+\frac{\pi}{6},\frac{23  \pi}{6}+\frac{\pi}{6},\frac{29\pi}{6}+\frac{\pi}{6  },\frac{35\pi}{6}+\frac{\pi}{6}
    x=\frac{6\pi}{36},\frac{12\pi}{36},\frac{18\pi}{36  },\frac{24\pi}{36},\frac{30\pi}{36},\frac{26\pi}{3  6}

    P.S
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I am having trouble solving the following equation:
    tan(3x-\frac{\pi}{6})
    range 0,2\pi
    this is what a have done, someone tell me where my mistake is.
    3x-\frac{\pi}{6}=\pi-\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi-\frac{\pi}{6},4\pi-\frac{\pi}{6},5\pi-\frac{\pi}{6},6\pi-\frac{\pi}{6}
    3x=\frac{5\pi}{6}+\frac{\pi}{6},\frac{11\pi}{6}+\f  rac{\pi}{6},\frac{17\pi}{6}+\frac{\pi}{6},\frac{23  \pi}{6}+\frac{\pi}{6},\frac{29\pi}{6}+\frac{\pi}{6  },\frac{35\pi}{6}+\frac{\pi}{6}
    x=\frac{6\pi}{36},\frac{12\pi}{36},\frac{18\pi}{36  },\frac{24\pi}{36},\frac{30\pi}{36},\frac{26\pi}{3  6}

    P.S
    HI

    That's not an equation , maybe you missed the = ??
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  3. #3
    Super Member
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I am having trouble solving the following equation:
    tan(3x-\frac{\pi}{6})
    range 0,2\pi
    this is what a have done, someone tell me where my mistake is.
    3x-\frac{\pi}{6}=\pi-\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi-\frac{\pi}{6},4\pi-\frac{\pi}{6},5\pi-\frac{\pi}{6},6\pi-\frac{\pi}{6}
    3x=\frac{5\pi}{6}+\frac{\pi}{6},\frac{11\pi}{6}+\f  rac{\pi}{6},\frac{17\pi}{6}+\frac{\pi}{6},\frac{23  \pi}{6}+\frac{\pi}{6},\frac{29\pi}{6}+\frac{\pi}{6  },\frac{35\pi}{6}+\frac{\pi}{6}
    x=\frac{6\pi}{36},\frac{12\pi}{36},\frac{18\pi}{36  },\frac{24\pi}{36},\frac{30\pi}{36},\frac{26\pi}{3  6}

    P.S
    It is meant to be: tan(3x-\frac{\pi}{6})=-1
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Paymemoney View Post
    It is meant to be: tan(3x-\frac{\pi}{6})=-1
    HI

    i will put it in degrees so that its easier for you to see but you will need to change it back to radian s.

    tan (3x-30)=-1

    3x-30 = 135 , 315 , 495 , 675 , 855 , 1035

    x=55 , 115 , 175 , 235 , 295 , 355

    the ones in red are the basic angles , the reference angle is 45 , since tan is -ve , it would be in the 2nd and 4th quadrant , then just add 360 for the subsequent angles . Just remember that

    0<x<360 , 0<3x<1080
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  5. #5
    Super Member 11rdc11's Avatar
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    New Orleans
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    Quote Originally Posted by Paymemoney View Post
    It is meant to be: tan(3x-\frac{\pi}{6})=-1
    Tan is -1 in the 2nd and 4th quad so

    3x-\frac{\pi}{6} = \frac{3\pi}{4}+2\pi n

    and

    3x-\frac{\pi}{6} = \frac{7\pi}{4} +2\pi n
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  6. #6
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    oh yeh wrong quadant
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