Question

решите пожалуйста 2sin^4x+3cos2x+1=0

Answer 1

2*(1-cos2x)²/4+3cos2x+1=0
(1-cos2x)²+6cos2x+2=0
1-2cos2x+cos²2x+6cos2x+2=0
cos²2x+4cos2x+3=0
a=cosx
a²+4a+3=0,a1+a2=-4 U a1*a2=3
a1=-1,cosx=-1⇒x=π+2πn
a2=-3,cosx=-3∉[-1;1]
x=π;3π

Answer 2

2sin^4x+3(cos^2x-sin^2x)+1=0 2sin^4x+3(1-sin^2x-sin^2x)+1=0 2sin^4x+3-6sin^2x+1=0 2sin^4x-6sin^2x+4=0 sin^2x=y