Question

Решите уравнение 8sin^2 2x+3sin 4x=7

Answer 1

$8sin^22x+3sin4x=7\\ \\ 8sin^22x+6sin2x*cos2x=7\\ \\ 8sin^22x+6sin2xcos2x-7(sin^22x+cos^22x)=0\\ \\ sin^22x+6sin2x*cos2x-7cos^22x=0|:cos^22x\\ \\ tg^22x+6tg2x-7=0\\ \\ tg2x=a\\ \\ a^2+6a-7=0\\ \\ D=6^2+4*7=64\\ \\ a_1=\frac{-6+8}{2}=1\\ \\ a_2=\frac{-6-8}{2}=-7\\ \\ tg2x=1\\ \\ 2x=\frac{\pi }{4}+\pi n, n \in Z\\ \\ x=\frac{\pi}{8}+\frac{\pi n}{2}, n \in Z\\ \\ tg2x=-7\\ \\ 2x=arcrg(-7)+ \pi n \in Z\\ \\ x=\frac{arctg(-7)}{2}+\frac{\pi n}{2}, n \in Z$