Question

Solving Trig Equations
a.) sec x + tan x=1

Answer 1

I don't know what to do.

Answer 2

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?

Answer 3

Would it be cot=1?

Answer 4

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

Answer 5

Do you add 2 cos x to the other side?

Answer 6

Then divide by 2

Answer 7

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?

Answer 8

2? Or is it cos?

Answer 9

Actually, it's both!

Answer 10

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

Answer 11

Add 1 to the other side?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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