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Trig Domain Question?
Find the domain of
F(x)= Sqrt(tan2x+pi)
So I made this sqrt(sin2x+pi/cos2x+pi)
And the domain would not exist where cos2x+pi=0
I'm not sure where to go from here?
Also, since its all square root x can't be negative?
2 Answers
- 6 years ago
If you wanted to rewrite tangent in sines and cosines, you wouldn't include the pi. What you want to do is take the inside of the square root, set it equal to zero, then solve for x. Then determine what x's make the inside negative and those numbers are what should be excluded from your domain. But to reorganize the inside, tan2x + pi = (cos2x/sin2x) + pi, then you'd set that equal to zero.
- 6 years ago
Did you mean this:
sqrt(tan(2x + pi))?
sqrt(t) exists when t >/= 0, so we need to define when tan(2x + pi) >/= 0
Well ,tan(m) >/= 0 when pi * k < m < pi/2 + pi * k, where k is an integer
(I'm going to let the "equals" part just be implied, rather than dig out the actual symbol for "greater than or equal to" and "lesser than or equal to")
pi * k < 2x + pi < pi/2 + pi * k
pi * k - pi < 2x < pi * k - pi/2
pi * (k - 1) < 2x < pi * (k - 1/2)
(pi/2) * (k - 1) < x < (pi/2) * (k - 1/2)
Just let k be an integer and that's your domain:
[(pi/2) * (k - 1) , (pi/2) * (k - 1/2)]
