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Problem Find the domain of the function h ( x ) = sec ( 4 x − π ) tan ( 2 x ) h(x)=\dfrac{\sec(4x-\pi)}{\tan(2x)} h ( x ) = tan ( 2 x ) sec ( 4 x − π ) .
a) R ∖ { x = π 4 + n π 2 , x = 3 π 8 + n π 4 , n ∈ Z } \mathbb{R}\setminus\left\{x=\dfrac{\pi}{4}+\dfrac{n\pi}{2},\,x=\dfrac{3\pi}{8}+\dfrac{n\pi}{4},\,n\in\mathbb{Z}\right\} R ∖ { x = 4 π + 2 nπ , x = 8 3 π + 4 nπ , n ∈ Z }
b) R ∖ { x = n π 2 , x = π 4 + n π 2 , x = 3 π 8 + n π 4 , n ∈ Z } \mathbb{R}\setminus\left\{x=\dfrac{n\pi}{2},\,x=\dfrac{\pi}{4}+\dfrac{n\pi}{2},\,x=\dfrac{3\pi}{8}+\dfrac{n\pi}{4},\,n\in\mathbb{Z}\right\} R ∖ { x = 2 nπ , x = 4 π + 2 nπ , x = 8 3 π + 4 nπ , n ∈ Z }
c) R ∖ { x = n π 2 , x = π 4 + n π 2 , x = π 8 + n π 4 , n ∈ Z } \mathbb{R}\setminus\left\{x=\dfrac{n\pi}{2},\,x=\dfrac{\pi}{4}+\dfrac{n\pi}{2},\,x=\dfrac{\pi}{8}+\dfrac{n\pi}{4},\,n\in\mathbb{Z}\right\} R ∖ { x = 2 nπ , x = 4 π + 2 nπ , x = 8 π + 4 nπ , n ∈ Z }
d) R ∖ { x = n π , x = π 4 + n π 2 , x = 3 π 8 + n π 4 , n ∈ Z } \mathbb{R}\setminus\left\{x=n\pi,\,x=\dfrac{\pi}{4}+\dfrac{n\pi}{2},\,x=\dfrac{3\pi}{8}+\dfrac{n\pi}{4},\,n\in\mathbb{Z}\right\} R ∖ { x = nπ , x = 4 π + 2 nπ , x = 8 3 π + 4 nπ , n ∈ Z }
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