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Math Help - Find the smallest positive

  1. #1
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    Find the smallest positive

    Find the smallest positive integer solution to
    \tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^  {\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}.
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  2. #2
    Senior Member JaneBennet's Avatar
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    \color{white}.\quad. \frac{\cos{96^\circ}+\sin{96^\circ}}{\cos{96^\circ  }-\sin{96^\circ}}

    =\ \frac{\sin{45^\circ}\cos{96^\circ}+\cos{45^\circ}\  sin{96^\circ}}{\cos{45^\circ}\cos{96^\circ}-\sin{45^\circ}\sin{96^\circ}}

    =\ \frac{\sin{(96^\circ+45^\circ)}}{\cos{(96^\circ+45  ^\circ)}}

    =\ \tan{141^\circ}

    =\ \tan{(141^\circ+180n^\circ)}\quad n\in\mathbb{Z}

    So find the smallest positive integer x such that 19x\equiv141\!\!\!\!\pmod{180}. The answer should be x = 159. (I worked it out using Excel. )
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  3. #3
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    Quote Originally Posted by JaneBennet View Post
    So find the smallest positive integer x such that 19x\equiv141\!\!\!\!\pmod{180}. The answer should be x = 159. (I worked it out using Excel. )[/SIZE][/FONT]
    Here is a non-excel solution. First 180 = 2^2\cdot 3^2\cdot 5.

    Thus, this is equivalent to,
    19x\equiv 141(\bmod 4)
    19x\equiv 141(\bmod 9)
    19x\equiv 141(\bmod 5)

    Which gets simplified to,
    -x\equiv 1(\bmod 4)
    x\equiv 6(\bmod 9)
    -x\equiv 1(\bmod 5)

    The first and third congruences combine since \gcd(4,5)=1.

    x\equiv -1(\bmod 20)
    x\equiv 6(\bmod 9)

    In first congruence subtract 20 once, in second congruence subtract 9 three times. To get*,

    x\equiv - 21(\bmod 20)
    x\equiv -21(\bmod 9)

    Thus, x\equiv -21(\bmod 180) \implies x\equiv 159(\bmod 180).


    *)You can use Chinese Remainder Theorem here, but I do not like to use that theorem, it is too computational. What you see here is my approach to combining modolus.
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