# Analytic Trigonometry

1. Dec 4, 2006

### Aviig

1.) sin2x=3cos2
2.) (3tan2x-1)(tan2x-3)=0
3.) cos3x=cosx
4.) 3tan3x=tanx

Those were a few of the problems in the section I missed on friday due to illness and I was wondering if anyone could walk me through them. Would be greatly appreciated!

They want to solve the equation for 0<=x<2pie for #1 and #2.
and
Find all solutions of the equation in the interval [0,2pie) algebraically for #3 and #4.

Oh and could anyone tell me how to make the symbols for pie, square root, and greater than or equal to for future reference?

2. Dec 4, 2006

(1) rewrite $$\sin^{2}x = 1-\cos^{2} x$$

(2) set each factor equal to 0

(3) Let $$u = \cos x$$. Then you have $$u^{3}-u = 0$$

(4) Again let $$u = \tan x$$. Then you have $$3u^{3}-u = 0$$

The symbols include: \pi, >, < , \leq, \geq, \sqrt{}

Last edited: Dec 4, 2006
3. Dec 4, 2006

### Aviig

Thanks a bunch, was a BIG help, I can already see myself loving this website, I have a mediocre teacher and havent been doing the best on tests, even though I ace quizzes, thanks again!

4. Dec 4, 2006

Here, regarding the LaTeX code reference: https://www.physicsforums.com/showthread.php?t=8997" (perhaps you already figured it out).

Last edited by a moderator: Apr 22, 2017
5. Dec 4, 2006

### Aviig

on 3 and 4 I am still having trouble, I can get 3 to:
cos3x=cosx
u3-u=0
cosx(cos2x-1)=0
cosx(sin2x)=0

then get stuck

and number 4 I cant figure past where you got me

6. Dec 4, 2006

For (3) $$\cos^{2}x-1 \;!= \sin^{2}x$$
set $$\cos x = 0$$ and $$\cos^{2}x-1 = 0$$
(4) Factor: $$u(3u^{2} -1) = 0$$, $$u = 0$$ and $$3u^{2} - 1 = 0$$