haleyking345 2 years ago Solving Trig Equations a.) sec x + tan x=1

1. haleyking345

I don't know what to do.

2. SithsAndGiggles

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

Would it be cot=1?

4. SithsAndGiggles

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

5. haleyking345

Do you add 2 cos x to the other side?

6. haleyking345

Then divide by 2

7. SithsAndGiggles

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

2? Or is it cos?

9. SithsAndGiggles

Actually, it's both!

10. SithsAndGiggles

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

11. haleyking345

Add 1 to the other side?

12. SithsAndGiggles

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

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

14. SithsAndGiggles

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

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

16. SithsAndGiggles

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