Question

2cos^2(x-3п/2)-sin(x-п) =0 [(5п/2) ; 4п ]

Answer 1

2sin²x+sinx=0
sinx(2sinx+1)=0
sinx=0⇒x=πn,n∈Z
5π/2≤πn≤4π
5/2≤n≤4
n=3⇒x=3π
n=4⇒x=4π
sinx=-1/2⇒x=(-1)^(k+1)*π/6+πk,k∈Z
k=2⇒x=-π/6+2π=11π/6∉[5π/2;4π]
k=3⇒x=π/6+3π=19π/6
k=4⇒x=-π/6+4π=23π/6
k=5⇒x=π/6+5π=31π/6∉[5π/2;4π]