Is this the question: \(\frac{\sin x}{\cos x} (2\cos x -1 )\)
Or
Is this the question: \(\frac{\sin x}{\cos x (2\cos x -1)}\)
The restrictions are all x values that cause the denominator to = 0.
So, set the denominator = 0 and solve for x.
\(\frac{\sin x}{\cos x (2\cos x -1)}\)
cos x (2 cos x - 1) = 0
Set each factor = 0
Either... cos x = 0 ...or... 2 cos x - 1 = 0
cos x = 0
x = arccos(0)
x = π/2 + π * n, where n is an integer
2 cos x - 1 = 0
2 cos x = 1
cos x = 1/2
x = arccos(1/2)
x = π/3 + 2π * n and x = 5π/3 + 2π * n, where n is an integer
So.. the restrictions are:
x ≠ π/2 + π * n
x ≠ π/3 + 2π * n
x ≠ 5π/3 + 2π * n , where n is an integer
*Thank you very much CPhill for the correction.*