What is the domain of y=-2tan(2θ)+1 ?

The domain of y=-2tan(2θ)+1 is pi/2 n<= θ< pi/4+pi/2 n\lor pi/4+pi/2 n<θ< pi/2+pi/2 n

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domain of y=-2tan(2θ)+1

domain

y = - 2 tan (2 θ) + 1

\frac{π}{2} n \leq θ < \frac{π}{4} + \frac{π}{2} n or \frac{π}{4} + \frac{π}{2} n < θ < \frac{π}{2} + \frac{π}{2} n

Interval Notation

[\frac{π}{2} n, \frac{π}{4} + \frac{π}{2} n) \cup (\frac{π}{4} + \frac{π}{2} n, \frac{π}{2} + \frac{π}{2} n)

Solution steps

Take the denominator(s) of

- 2 tan (2 θ) + 1

and compare to zero:

θ = \frac{π}{4} + \frac{π}{2} n

Periodicity of

- 2 tan (2 θ) + 1 : \frac{π}{2}

Combine period:

\frac{π}{2} n \leq θ < \frac{π}{2} + \frac{π}{2} n

and undefined points.

The domain is

\frac{π}{2} n \leq θ < \frac{π}{4} + \frac{π}{2} n or \frac{π}{4} + \frac{π}{2} n < θ < \frac{π}{2} + \frac{π}{2} n

f (x) = \frac{4 x - 5}{x + 6}

f (x) = \frac{2 x}{x ^{2}}

(\frac{5}{6})^{\frac{x ^{2} - 3 x}{x ^{2}}}

lo g_{10} (5 - 2 x)

\frac{2}{x ^{2} - 1} + x + 3

What is the domain of y=-2tan(2θ)+1 ?
The domain of y=-2tan(2θ)+1 is pi/2 n<= θ< pi/4+pi/2 n\lor pi/4+pi/2 n<θ< pi/2+pi/2 n