Question

cos^2x-sin^2x+sinx=0

Answer 1

cos²x-sin²x+sinx=0
cos2x+sinx=0
1-2sin²x+sinx=0
sinx=a
2a²-a-1=0
D=1+8=9
a1=(1-3)/4=-1/2⇒sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈z
a2=(1+3)/4=1⇒sinx=1⇒x=π/2+2πn,n∈z