Question

Решить уравнение 1+2|sinx|=2cos2x

Answer 1

1+2sinx=2cos2x

1+2sinx=2(1-2sin^2x)

1+2sinx=2-4sin^2x

4sin^2x+2sinx-1=0

sinx=(-1+-sqrt(5))/4  sinx>=0

sinx=(sqrt(5)-1)/4

x=(-1)^k*arcsin(sqrt(5)-1)/4)+Пk

 

sinx<0

1-2sinx=2-4sin^2x

4sin^2x-2sinx-1=0

x=(1+-sqrt(5))/4

x=1-sqrt(5))/4

 

x=(-1)^larcsin((1-sqrt(5)/4)+Пk