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f(x)= $\sqrt{(tan(2x+\pi)}$
Allright, so i know you cannot have a number less than zero under the square root sign and that tangent cannot equal π/2+nπ. So should i try to find the domain of the tan function and see where it is greater than zero?
If so, how would i go about doing that?
So I tried to solve 0 ≤ 2x + π < π / 2 to get to A ≤ x < B but I am stuck...
First as a minor point, I don't think $f$ is considered a trig function, as there is also a root involved. Anyway, now you can work from the outside in to figure out which conditions are necessary on $x$ so that $f$ is valued. Since everything under the root must be greater than or equal to zero, we know $$\tan(2x+\pi) \geq 0$$ and if you think about the unit circle, you know tangent is positive in quadrants $I$ and $III$ (But not defined at $\pm \frac{\pi}{2}$. So you want $$0 \leq 2x+ \pi<\frac{\pi}{2} \quad \text{and} \quad \pi \leq 2x+\pi < \frac{3\pi}{2}$$ For the first inequality, subtract $\pi$ and divide by $2$ to get $$-\frac{\pi}{2} \leq x <-\frac{\pi}{4}$$ repeat for the second inequality. Next recall that if you begin at an angle and travel $2\pi$ radians, you'll end up back at the same angle, meaning that you can generalize the fist inequality to be $$-\frac{\pi}{2}+2n\pi \leq x <-\frac{\pi}{4}+2n\pi$$ and a similar generalization can be made about the second inequality. We can then define a union $$D_1 = \bigcup_{n \in \Bbb{Z}}\left[-\frac{\pi}{2}+2n\pi,-\frac{\pi}{4}+2n\pi \right)$$ obtained from the generalized inequality above. Repeat by letting $D_2$ be the union obtained from the generalization of the second inequality. Then your final answer (domain of $f$) is simply $D_1 \cup D_2$.
