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haleyking345
Solving Trig Equations a.) sec x + tan x=1
I don't know what to do.
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 \[\sin^2x+\cos^2x=1\] \[1+\sqrt{1-\cos^2x}=\cos x\\ \sqrt{1-\cos^2x}=\cos x-1\\ 1-\cos^2x=(\cos x-1)^2\\ 1-\cos^2x=\cos^2x-2\cos x+1\] Does that help?
Would it be cot=1?
Well, that last equation simplifies to \[2\cos^2x-2\cos x=0\] What can you do with that?
Do you add 2 cos x to the other side?
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?
Actually, it's both!
Factor out a 2cosx from both terms, and you get \[2\cos x \;(\cos x - 1)=0\] What's next?
Add 1 to the other side?
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 \[2\cos x=0 \text{ or } \cos x-1=0\]
Oh so whhen they are in paranthesis, you can set them equal to 0?
Parentheses aren't exactly the qualifying factor here, no. When you have the product of two things equal to 0, such as \[(ax+b)(cx+d)=0, \text{ then you can split up the left side and get}\\ ax+b=0 \text{ AND }cx+d=0.\\ \text{Solving both equations gives you $x=-\frac{b}{a}$ and $x=-\frac{c}{d}$ as solutions.}\]
The parentheses are there to show that the term (cosx - 1) is a factor of 0, as is 2cosx.
There's a big difference between \[2\cos x\;(\cos x-1) \text{ and } 2\cos x \cos x -1\]