Explain how many solutions the following trigonometric equation should have:
2cos^2x - 1 = 0, when 0 ā¤ x ā¤ 2š
2cos^2x - 1 = cos(2x)
so 2x = (š/2 or 3š/2) + k*2š
x = (š/4 or 3š/4) + kš
so how many values of k produce a result in [0,2š]?