Find all real x such that 0 < x < π and 8/(3 sin x - sin 3x) + 3 sin2x ≤ 5.
Answer
π/2
Solution
We have 3 sin x - sin 3x = 4 sin3x. Put s = sin x. Then we want 2/s3 + 3s2 ≤ 5. Note that since 0 < x < π we have s positive. But by AM/GM we have 1/s3 + 1/s3 + s2 + s2 + s2 > 5 with equality iff s = 1, so we must have sin x = 1 and hence x = π/2.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
7 March 2004
Last corrected/updated 7 Mar 04