103 Trigonometry Problems

5. Solutions to Advanced Problems 129
Solution: By the arithmetic–geometric means inequality,
cos 3 x + cos x
4
≥ cos 2 x
for x such that cos x ≥ 0. Because triangle ABC is acute, cos A, cos B, and
cos C are nonnegative. Setting x = A, x = B, x = C and adding the resulting
inequalities yields
x 1 + x 3 ≥ cos 2 A + cos 2 B + cos 2 C + 2 cos A cos B cos C = 2x 2 .
Consequently,
x 1 + x 2 + x 3 ≥ 3x 2 = 3 2 ,
by Introductory Problem 24(d).
6. Find the sum of all x in the interval [0, 2π] such that
3 cot 2 x + 8 cot x + 3 = 0.
Solution: Consider the quadratic equation
3u 2 + 8u + 3 = 0.
The roots of the above equation are u 1 = −8+2√ 7
6
and u 2 = −8−2√ 7
6
. Both
roots are real, and their product u 1 u 2 is equal to −1 (by Viète’s theorem).
Because y = cot x is a bijection from the interval (0,π)to the real numbers,
there is a unique pair of numbers x 1,1 and x 2,1 with 0 < x 1,1 ,x 2,1 < π
such that cot x 1,1 = u 1 and cot x 2,1 = u 2 . Because u 1 ,u 2 are negative, π 2 <
x 1,1 ,x 2,1