The number of real solutions of the equation sin(∞∑i=1xi + 1 - x∞∑i=1(x²)i) = π/2 - cos(∞∑i=1(-x²)i - ∞∑i=1(-x)i) lying in the interval (-1/2, 1/2) is

∞∑i=1xi + 1 = x/(1 - x)
∞∑i=1(x²)i = x²/(1 - x²)
∞∑i=1(-x²)i = -x²/(1 + x²)
∞∑i=1(-x)i = -x/(1 + x)
To have real solutions
∞∑i=1xi + 1 - x∞∑i=1(x²)i = ∞∑i=1(-x²)i - ∞∑i=1(-x)i
x/(1 - x) - x²/(1 - x²) = -x²/(1 + x²) + x/(1 + x)
x(x³ + 2x² + 5x - 1) = 0
∴ x = 0 and let f(x) = x³ + 2x² + 5x - 1
f(1/2) * f(-1/2) < 0
Hence two solutions exist.