Problem:
Suppose x is in the interval [0,π/2] and log24sinx(24cosx)=23. Find 24cot2x.
Solution:
By the change of base formula, log24sinx(24cosx)=log10(24sinx)log10(24cosx). The given equation then becomes 2log10(24cosx)=3log10(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+sin2x−1=0. The left side of this equation can be rewritten as
which equals (3sinx−1)(8sin2x+3sinx+1). Thus the solutions of the original equation are sinx=1/3 and two non-real complex values. Then cosx=38 and cotx=8. The required result is 24⋅8=192.