2cos^2пx+sin п x-1=0

Question

2cos^2пx+sin п x-1=0

Answer 1

2(1-sin²πx)+sinπx-1=0
-2sin²πx +sinπx+1=0
z=sinπx
-2z²+z+1=0
2z²-z-1=0   D=1+8=9  √D=3 
z1=1/4[1+3]=1    z2=1/4[1-3]=-1/2
sinπx=1   πx=π/2+2πk   x=1/2+2k   k∈Z
sinπx=-1/2   πx=(-1)ⁿarcsin(-1/2)+πk=(-1)ⁿ⁺¹π/6+πk
                       x=(-1)ⁿ⁺¹/6+k   k∈Z