#1**+2 **

Is this the question: \(\frac{\sin x}{\cos x} (2\cos x -1 )\)

Or

Is this the question: \(\frac{\sin x}{\cos x (2\cos x -1)}\)

hectictar
Apr 13, 2017

#3**+2 **

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.*

hectictar
Apr 13, 2017