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ella [17]
3 years ago
6

Which of the following describes the domain of y = tan x, where n is any integer? A.x≠2nπ B. x≠π/2+nπ C. x≠nπ D.x≠nπ/2

Mathematics
1 answer:
lidiya [134]3 years ago
8 0
\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}

therefore, that fraction will become undefined if the denominator ever turns to 0.

now, recall that cos(π/2) is 0, and cos(3π/2) is 0, and cos(5π/2) is also zero and so on, thus

\bf tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{cos\left( \frac{\pi }{2} \right)}\implies tan\left( \frac{\pi }{2} \right)=\cfrac{sin\left( \frac{\pi }{2} \right)}{0}
\\\\\\
tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{cos\left( \frac{3\pi }{2} \right)}\implies tan\left( \frac{3\pi }{2} \right)=\cfrac{sin\left( \frac{3\pi }{2} \right)}{0}


\bf tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{cos\left( \frac{5\pi }{2} \right)}\implies tan\left( \frac{5\pi }{2} \right)=\cfrac{sin\left( \frac{5\pi }{2} \right)}{0}
\\\\\\
tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{cos\left( \frac{7\pi }{2} \right)}\implies tan\left( \frac{7\pi }{2} \right)=\cfrac{sin\left( \frac{7\pi }{2} \right)}{0}

\bf tan\left( \frac{\pi }{2}+\pi  \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi  \right)}{cos\left( \frac{\pi }{2}+\pi  \right)}\implies tan\left( \frac{\pi }{2}+\pi  \right)=\cfrac{sin\left( \frac{\pi }{2}+\pi  \right)}{0}
\\\\\\
domain\implies \{x|x\in \mathbb{R}, ~~x\ne \frac{\pi }{2}+n\pi,~~n\in \mathbb{Z} \}
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Please find the image file of the graph.

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Solve the parenthesis of each term.

=7\left\sqrt[3]{2x}\right-3\left\sqrt[3]{16x}\right-3\left\sqrt[3]{8x}\right

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So, option c is correct

5 0
3 years ago
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