2nd VMO 1963

Solve the equation sin³x cos 3x + cos³x sin 3x = 3/8.

Answer

7½^o + k90^o or 37½^o + k90^o

Solution

We have sin 3x = 3 sin x - 4 sin³x, cos 3x = 4 cos³x - 3 cos x. So we need 4 sin³x cos³x - 3 sin³x cos x + 3 sin x cos³x - 4 sin³x cos³x = 3/8 or 8 sin x cos x(cos²x - sin²x) = 1, or 4 sin 2x cos 2x = 1 or sin 4x = 1/2. Hence 4x = 30^o + k360^o or 150^o + k360^o. So x = 7½^o + k90^o or 37½^o + k90^o.

Thanks to Suat Namli

2nd VMO 1963

© John Scholes
jscholes@kalva.demon.co.uk
7 March 2004
Last corrected 7 Mar 04