0,\; sinx\ne 0\\\\sinx=\frac{1}{2}\\\\x_2=(-1)^{k}\frac{\pi}{6}+\pi k,k\in Z" alt="8sinxcosx=\frac{cosx}{sin^2x}\\\\cosx(8sinx-\frac{1}{sin^2x})=0\\\\1.\quad cosx=0,\; x_1=\frac{\pi}{2}+\pi n,n\in Z\\\\2.\quad \frac{8sin^3x-1}{sin^2x}=0\\\\\frac{(2sinx-1)(4sin^2x+2sinx+1)}{sin^2x}=0\\\\2sinx-1=0,\; 4sin^2x+2sinx+1>0,\; sinx\ne 0\\\\sinx=\frac{1}{2}\\\\x_2=(-1)^{k}\frac{\pi}{6}+\pi k,k\in Z" align="absmiddle" class="latex-formula">