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haleyking345 Group TitleBest ResponseYou've already chosen the best response.0
I don't know what to do.
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Recall\[\sec x=\frac{1}{\cos x}\\\tan x = \frac{\sin x}{\cos x}\] \[\sec x+\tan x=1\\ \frac{1}{\cos x}+\frac{\sin x}{\cos x}=1\\ \frac{1+\sin x}{\cos x}=1\\ 1+\sin x=\cos x\] 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?
 one year ago

haleyking345 Group TitleBest ResponseYou've already chosen the best response.0
Would it be cot=1?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Well, that last equation simplifies to \[2\cos^2x2\cos x=0\] What can you do with that?
 one year ago

haleyking345 Group TitleBest ResponseYou've already chosen the best response.0
Do you add 2 cos x to the other side?
 one year ago

haleyking345 Group TitleBest ResponseYou've already chosen the best response.0
Then divide by 2
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Sure, 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?
 one year ago

haleyking345 Group TitleBest ResponseYou've already chosen the best response.0
2? Or is it cos?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Actually, it's both!
 one year ago

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

haleyking345 Group TitleBest ResponseYou've already chosen the best response.0
Add 1 to the other side?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
No, 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 \[2\cos x=0 \text{ or } \cos x1=0\]
 one year ago

haleyking345 Group TitleBest ResponseYou've already chosen the best response.0
Oh so whhen they are in paranthesis, you can set them equal to 0?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Parentheses aren't exactly the qualifying factor here, no. When you have the product of two things equal to 0, such as \[(ax+b)(cx+d)=0, \text{ then you can split up the left side and get}\\ ax+b=0 \text{ AND }cx+d=0.\\ \text{Solving both equations gives you $x=\frac{b}{a}$ and $x=\frac{c}{d}$ as solutions.}\]
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
The parentheses are there to show that the term (cosx  1) is a factor of 0, as is 2cosx.
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
There's a big difference between \[2\cos x\;(\cos x1) \text{ and } 2\cos x \cos x 1\]
 one year ago
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