Solve 2cos^2x - 1 = 0 for 0° < x < 360°

Answer 1

Solve 2cos^2x - 1 = 0 for 0° < x < 360°

I asssume this means cos²(x)

2cos²(x) - 1 = 0

cos²(x) =1/2

cos(x) = ±√(1/2) = ±(1/2)*√2

x = ±pi/4, ±3pi/4