Mastering Trigonometry Equations: Solving for Missing Solutions in [0,2π)

[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?

 

[courtrigrad]

(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{}

 

[Aviig]

Thanks a bunch, was a BIG help, I can already see myself loving this website, I have a mediocre teacher and haven't been doing the best on tests, even though I ace quizzes, thanks again!

 

[radou]

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

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

 

[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 can't figure past where you got me

 

[courtrigrad]

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