Problem:
Suppose x is in the interval [0,Ο/2] and log24sinxβ(24cosx)=23β. Find 24cot2x.
Solution:
By the change of base formula, log24sinxβ(24cosx)=log10β(24sinx)log10β(24cosx)β. The given equation then becomes 2log10β(24cosx)=3log10β(24sinx). This equation is equivalent to 242cos2x=243sin3x, which, after dividing by 242 and using the Pythagorean identity for the cosine function yields the equation 24sin3x+sin2xβ1=0. The left side of this equation can be rewritten as
which equals (3sinxβ1)(8sin2x+3sinx+1). Thus the solutions of the original equation are sinx=1/3 and two non-real complex values. Then cosx=38ββ and cotx=8β. The required result is 24β 8=192β.