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anonymous
 3 years ago
Solving Trig Equations
a.) sec x + tan x=1
anonymous
 3 years ago
Solving Trig Equations a.) sec x + tan x=1

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't know what to do.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Recall\[\sec x=\frac{1}{\cos x}\\\tan x = \frac{\sin x}{\cos x}\] \[\sec x+\tan x=1\\ \frac{1}{\cos x}+\frac{\sin x}{\cos x}=1\\ \frac{1+\sin x}{\cos x}=1\\ 1+\sin x=\cos x\] Use the identity \[\sin^2x+\cos^2x=1\] \[1+\sqrt{1\cos^2x}=\cos x\\ \sqrt{1\cos^2x}=\cos x1\\ 1\cos^2x=(\cos x1)^2\\ 1\cos^2x=\cos^2x2\cos x+1\] Does that help?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, that last equation simplifies to \[2\cos^2x2\cos x=0\] What can you do with that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you add 2 cos x to the other side?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sure, you could do that, but then you get \[\cos^2x=\cos x\] Then your first instinct would be to divide both sides by a cosx, which would eliminate a potential solution. Here's a hint: What's a common factor of 2cos²x and 2cosx?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Factor out a 2cosx from both terms, and you get \[2\cos x \;(\cos x  1)=0\] What's next?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Add 1 to the other side?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, that's not it. Think of it this way: When you have \[a\cdot b = 0, \text{ either $a$ or $b$ (or both) must be zero, right?}\] This means that for 2cosx (cosx  1) to be equal to 0, either \[2\cos x=0 \text{ or } \cos x1=0\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh so whhen they are in paranthesis, you can set them equal to 0?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Parentheses aren't exactly the qualifying factor here, no. When you have the product of two things equal to 0, such as \[(ax+b)(cx+d)=0, \text{ then you can split up the left side and get}\\ ax+b=0 \text{ AND }cx+d=0.\\ \text{Solving both equations gives you $x=\frac{b}{a}$ and $x=\frac{c}{d}$ as solutions.}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The parentheses are there to show that the term (cosx  1) is a factor of 0, as is 2cosx.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There's a big difference between \[2\cos x\;(\cos x1) \text{ and } 2\cos x \cos x 1\]
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