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330
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determine all restrictions

sin x/cos x (2cosx-1)

nlbradshaw0430  Apr 13, 2017
 #1
avatar+7153 
+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
 #2
avatar+36 
+1

the second one

nlbradshaw0430  Apr 13, 2017
 #3
avatar+7153 
+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
edited by hectictar  Apr 13, 2017

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