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Thread: Find the smallest positive

  1. #1
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    Find the smallest positive

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

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

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

    $\displaystyle =\ \tan{141^\circ}$

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

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

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

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

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

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

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

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

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


    *)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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