A problem by Rohan Chandra

Level pending

Find the smallest positive integer solution to tan ⁡ 19 x 0 \tan 19x^{0} = cos ⁡ 9 6 0 + sin ⁡ 9 6 0 cos ⁡ 9 6 0 − sin ⁡ 9 6 0 \frac{\cos 96^{0}+\sin 96^{0}}{\cos 96^{0}-\sin 96^{0}}


The answer is 159.

Rohan Chandra by Rohan Chandra , 24, India

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1 solution

Rohan Chandra
Oct 1, 2014

cos ⁡ 9 6 0 + s i n 9 6 0 cos ⁡ 9 6 0 − s i n 9 6 0 \frac{\cos 96^{0}+sin 96^{0}}{\cos 96^{0}-sin 96^{0}} = s i n 18 6 0 + s i n 9 6 0 s i n 18 6 0 − s i n 9 6 0 \frac{sin 186^{0}+sin 96^{0}}{sin 186^{0}-sin 96^{0}} = s i n ( 14 1 0 + 4 5 0 ) + s i n ( 14 1 0 − 4 5 0 ) s i n ( 14 1 0 + 4 5 0 ) − s i n ( 14 1 0 − 4 5 0 ) \frac{sin (141^{0}+45^{0})+sin (141^{0}-45^{0})}{sin (141^{0}+45^{0})-sin (141^{0}-45^{0})} = 2 s i n 14 1 0 c o s 4 5 0 2 c o s 14 1 0 s i n 4 5 0 \frac{2sin141^{0}cos45^{0}}{2cos141^{0}sin45^{0}} = tan ⁡ 14 1 0 \tan 141^{0}

Since the period of the tan function is 18 0 0 180^{0} & and the tangent function is one-to-one over each period of its domain.

therefore, 19x = 141 (mod 180) since, 1 9 2 19^{2} = 361 = 1 (mod 180) , multiplying both sides by 19 yields x = 141 × 19 141 \times 19 = ( 140 + 1 ) × ( 18 + 1 ) (140+1) \times (18+1) = 0 + 140 + 18 + 1 = 159 (mod 180).

Hence, the smallest positive solution of x is = 159 \boxed{159}

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