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# determine all restrictions

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determine all restrictions

sin x/cos x (2cosx-1)

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#1
+5589
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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
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the second one

#3
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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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