\sin x=\pm1=>8+0+0-4 \neq 0;\\
b)\cos x \neq 0\\
\frac{8\sin^2x+\cos x\sin x+\cos^2x-4}{\cos^2x}=0;\\
8tg^2x+tgx+1- \frac{4}{\cos^2x}=0;\\
| \frac{1}{\cos^2x}=\frac{\sin^2x+\cos^x}{\cos^2x}=1+tg^2x|\\
8tg^2x+tgx+1- 4-4tg^2x=0\\
4tg^2x+tgx-3=0; \ \ t=tgx;\\
4t^2+t-3=0;\\
D=b^2-4ac=1+4\cdot3\cdot4=1+48=49=7^2;\\
" alt="8\sin^2x+\cos x\sin x+\cos^2x-4=0;\\
a) \cos x=0=>\sin x=\pm1=>8+0+0-4 \neq 0;\\
b)\cos x \neq 0\\
\frac{8\sin^2x+\cos x\sin x+\cos^2x-4}{\cos^2x}=0;\\
8tg^2x+tgx+1- \frac{4}{\cos^2x}=0;\\
| \frac{1}{\cos^2x}=\frac{\sin^2x+\cos^x}{\cos^2x}=1+tg^2x|\\
8tg^2x+tgx+1- 4-4tg^2x=0\\
4tg^2x+tgx-3=0; \ \ t=tgx;\\
4t^2+t-3=0;\\
D=b^2-4ac=1+4\cdot3\cdot4=1+48=49=7^2;\\
" align="absmiddle" class="latex-formula">
tgx_{1}=-1; x_{1}= \frac{3\pi}{4}+\pi n, n\in Z\\
t_{1}= \frac{-b+\sqrt{D}}{2\cdota}= \frac{-1+7}{8}= \frac{3}{4} ;=>tgx_{2}=-1; x_{2}= arctg \frac{3}{4} +\pi n, n\in Z\\ " alt="t_{1}= \frac{-b-\sqrt{D}}{2\cdota}= \frac{-1-7}{8}=-1;=>tgx_{1}=-1; x_{1}= \frac{3\pi}{4}+\pi n, n\in Z\\
t_{1}= \frac{-b+\sqrt{D}}{2\cdota}= \frac{-1+7}{8}= \frac{3}{4} ;=>tgx_{2}=-1; x_{2}= arctg \frac{3}{4} +\pi n, n\in Z\\ " align="absmiddle" class="latex-formula">