2cos^2 пx+sin п x-1=0

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