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

1. anonymous

I don't know what to do.

2. anonymous

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?

3. anonymous

Would it be cot=1?

4. anonymous

Well, that last equation simplifies to $2\cos^2x-2\cos x=0$ What can you do with that?

5. anonymous

Do you add 2 cos x to the other side?

6. anonymous

Then divide by 2

7. anonymous

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?

8. anonymous

2? Or is it cos?

9. anonymous

Actually, it's both!

10. anonymous

Factor out a 2cosx from both terms, and you get $2\cos x \;(\cos x - 1)=0$ What's next?

11. anonymous

Add 1 to the other side?

12. anonymous

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 $2\cos x=0 \text{ or } \cos x-1=0$

13. anonymous

Oh so whhen they are in paranthesis, you can set them equal to 0?

14. anonymous

Parentheses aren't exactly the qualifying factor here, no. When you have the product of two things equal to 0, such as $(ax+b)(cx+d)=0, \text{ then you can split up the left side and get}\\ ax+b=0 \text{ AND }cx+d=0.\\ \text{Solving both equations gives you x=-\frac{b}{a} and x=-\frac{c}{d} as solutions.}$

15. anonymous

The parentheses are there to show that the term (cosx - 1) is a factor of 0, as is 2cosx.

16. anonymous

There's a big difference between $2\cos x\;(\cos x-1) \text{ and } 2\cos x \cos x -1$