Тригонометрия #39 В ответе должен быть 3)

Решите задачу:

$\displaystyle \frac{sin^22x-cos^23x}{cos5x}=\frac{(2sinx*cosx)^2-(4cos^3x-3cosx)^2}{cos(2x+3x)}=\\\\=\frac{4sin^2x*cos^2x-(16cos^6x-24cos^4x+9cos^2x)}{cos3x*cos2x-sin3x*sin2x}=\\\\=\frac{cos^2x(4(1-cos^2x)-16cos^4x+24cos^2x-9)}{(4cos^2x-3cosx)(2cos^2x-1)-(3sinx-4sin^3x)(2sinx*cosx)}=\\\\=\frac{cos^2x(20cos^2x-16cos^4x-5)}{16cos^5x-20cos^3x+5cosx}=\frac{cos^2x(20cos^2x-16cos^4x-5)}{-cosx(20cos^2x-16cos^4x-5)}=\\\\=\frac{cos^2x}{-cosx}=-cosx$