Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.
Find the smallest positive integer solution to \(tan19x=\frac{cos96+sin96}{cos96-sin96}\).
Functions
Trigonometry
Integers
Answer: is 159.
AIME I, 1996, Question 10
Plane Trigonometry by Loney
\(\frac{cos96+sin96}{cos96-sin96}\)
=\(\frac{sin(90+96)+sin96}{sin(90+96)-sin96}\)
=\(\frac{sin186+sin96}{sin186-sin96}\)
=\(\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}\)
=\(\frac{2sin141cos45}{2cos141sin45}\)
=tan141
here \(tan(180+\theta)\)=\(tan\theta\)
\(\Rightarrow 19x=141+180n\) for some integer n is first equation
multiplying equation with 19 gives
\(x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)\) [since 2679 divided by 180 gives remainder 159]
\(\Rightarrow x=159\).