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Calculus/q
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Question
How do I show that
sinx-secx=0 has no solutions?
Also how do I solve sinx=cosx
It is known that |sin(x)| <= 1.
It is known that |sec(x)| >= 1.
It is also known the sec(x) = 1/cos(x).
Now sin(x) is 1 at (2n+1)π where n is an integer
and sec(x) is 1 at 2mπ where m is an integer.
Now basically, we know that 2n+1 is always odd
and 2m is always even. Odds and evens are never
equal, so the sin(x) and sec(x) are never both
1 at the same point, therefore the difference
can never be 0.
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Answers by Expert:
Scotto
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Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.
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