Find the smallest positive

**perash** · March 26th 2008, 09:30 AM

Find the smallest positive integer solution to
$\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^ {\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ .

**JaneBennet** · March 26th 2008, 10:16 AM

$\color{white}.\quad.$ $\frac{\cos{96^\circ}+\sin{96^\circ}}{\cos{96^\circ }-\sin{96^\circ}}$

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

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

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

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

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.

)

**ThePerfectHacker** · March 26th 2008, 05:32 PM

Originally Posted by JaneBennet

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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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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