Question

sin2x+sin6x=6sin^2(2x)​

Answer 1

sin2x=2sinx\cdot cosx\\
6sin^2x+2sinx\cdot cosx=2\\
6sin^2x+2sinx\cdot cosx-2(sin^2x+cos^2x)=0\\
4sin^2x+2sinx\cdot cosx-2cos^2x=0\\
 \frac{4sin^2x+2sinx\cdot cosx-2cos^2x}{cos^2x}=0\\
4tg^2x+2tgx-2=0\\
2tg^2x+tgx-1=0\\
2tg^2x+tgx-1=0\\
2tg^2x+2tgx-tgx-1=0\\
2tgx(1+tgx)-(tgx+1)=0\\
(2tgx-1)(tgx+1)=0\\
tgx_1=0,5\\ tgx_2=-1\\
x_1=arctg0,5+ \pi n\\
x_2=- \frac{ 3\pi }{4} + \pi n