## haleyking345 Group Title Solving Trig Equations a.) sec x + tan x=1 one year ago one year ago

1. haleyking345 Group Title

I don't know what to do.

2. SithsAndGiggles Group Title

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. haleyking345 Group Title

Would it be cot=1?

4. SithsAndGiggles Group Title

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

5. haleyking345 Group Title

Do you add 2 cos x to the other side?

6. haleyking345 Group Title

Then divide by 2

7. SithsAndGiggles Group Title

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. haleyking345 Group Title

2? Or is it cos?

9. SithsAndGiggles Group Title

Actually, it's both!

10. SithsAndGiggles Group Title

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

11. haleyking345 Group Title

Add 1 to the other side?

12. SithsAndGiggles Group Title

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. haleyking345 Group Title

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

14. SithsAndGiggles Group Title

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. SithsAndGiggles Group Title

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

16. SithsAndGiggles Group Title

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