Sin^2*x+9cos^2*x=5sin2x

Question

Sin^2*x+9cos^2*x=5sin2x

Answer 1

sin^2*x+9cos^2*x=5(2sinx*cosx)
sin^2*x+9cos^2*x-10sinx*cosx=0

(sin^2*x)/cos^2*x+(9cos^2*x)/cos^2*x-(10sinx*cosx)/cos^2*x=0
tg^2*x+9-10tgx=0
Пусть tgx=t
t^2+9-10t=0
D= 100-36=64
x 1 и 2 =(10+-8)/2=9 и 1
tgx=9     and     tgx=1
x= arctg9+пиk              and         x=пи/4+пиN