Explain how many solutions the following trigonometric equation should have:

2cos^2x - 1 = 0, when 0 ā‰¤ x ā‰¤ 2š›‘

1 answer

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›‘]?