A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

haleyking345
 one year ago
Best ResponseYou've already chosen the best response.0I don't know what to do.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Recall\[\sec x=\frac{1}{\cos x}\\\tan x = \frac{\sin x}{\cos x}\] \[\sec x+\tan x=1\\ \frac{1}{\cos x}+\frac{\sin x}{\cos x}=1\\ \frac{1+\sin x}{\cos x}=1\\ 1+\sin x=\cos x\] Use the identity \[\sin^2x+\cos^2x=1\] \[1+\sqrt{1\cos^2x}=\cos x\\ \sqrt{1\cos^2x}=\cos x1\\ 1\cos^2x=(\cos x1)^2\\ 1\cos^2x=\cos^2x2\cos x+1\] Does that help?

haleyking345
 one year ago
Best ResponseYou've already chosen the best response.0Would it be cot=1?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Well, that last equation simplifies to \[2\cos^2x2\cos x=0\] What can you do with that?

haleyking345
 one year ago
Best ResponseYou've already chosen the best response.0Do you add 2 cos x to the other side?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Sure, 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?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Actually, it's both!

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Factor out a 2cosx from both terms, and you get \[2\cos x \;(\cos x  1)=0\] What's next?

haleyking345
 one year ago
Best ResponseYou've already chosen the best response.0Add 1 to the other side?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1No, 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 \[2\cos x=0 \text{ or } \cos x1=0\]

haleyking345
 one year ago
Best ResponseYou've already chosen the best response.0Oh so whhen they are in paranthesis, you can set them equal to 0?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Parentheses aren't exactly the qualifying factor here, no. When you have the product of two things equal to 0, such as \[(ax+b)(cx+d)=0, \text{ then you can split up the left side and get}\\ ax+b=0 \text{ AND }cx+d=0.\\ \text{Solving both equations gives you $x=\frac{b}{a}$ and $x=\frac{c}{d}$ as solutions.}\]

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1The parentheses are there to show that the term (cosx  1) is a factor of 0, as is 2cosx.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1There's a big difference between \[2\cos x\;(\cos x1) \text{ and } 2\cos x \cos x 1\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.