1-sinx/2=cosx
1-sinx/2-1+2sin²x/2=0
2sinx/2-sinx/2=0sinx/2(2sinx/2-1)=0
sinx/2=0⇒x/2=πn⇒x=2πn,n∈z
sinx/2=1/2⇒x/2=(-1)^n*π/6+πn⇒x=(-1)^n*π/3+2πn,n∈z
1+4cosx=cos2x
2cos²x-1-4cosx-1=0
2cos²x-4cosx-2=0
cos²x-2cosx-1=0
cosx=a
a²-2a-1=0
D=4+4=8
a1=(2-2√2)/2=1-√2⇒cosx=1-√2⇒x=+-(π-arccos(√2-1))+2πn,n∈z
a2=1+√2⇒cosx=1+√2>1 нет решения
1) ( 16 ^ n - 2 ) * ( 2 ^ 2n - 3 ) = ( ( 2 ^ 4 ) ^ n - 2 ) * ( 2 ^ 2n - 3 ) = ( 2 ^ 4n - 8 ) * ( 2 ^ 2n - 3 ) = 2 ^ ( 6n - 11 )
2) 8 ^ ( 2n - 4 ) = ( 2 ^ 3 ) ^ ( 2n - 4 ) = 2 ^ ( 6n - 12 )
3) ( 2 ^ 6n - 11 ) : ( 2 ^ 6n - 12 ) = 2 ^ ( 6n - 11 - 6n + 12 ) = 2 ^ 1 = 2
Ответ 2