Question

cos^2x/2-sin^2x/2=sin(п/2-2x)
пожалуйста решите

Answer 1

cosx=cos2x
2cosx-1-cosx=0
cosx=a
2a²-a-1=0
D=1+8=9
a1=(1-3)/4=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn,n∈z
a2=(1+3)/4=1⇒cosx=1⇒x=2πk,k∈z