# [SOLVED] Trig

• May 6th 2008, 04:05 PM
>_<SHY_GUY>_<
[SOLVED] Trig
can some one please show me how to show that this equation is true?

cot^2x-cos^2x= cot^2x * cos^2x

and i have one more question:
in solving:
4cos^2x-3=0
i get +- sqrt. 3/2
this means that there are 4 answers, but in the book, it gives me 2 :
pi/6+npi and 5pi/6+npi
why?

thanks
• May 6th 2008, 04:13 PM
Jhevon
Quote:

Originally Posted by >_<SHY_GUY>_<
can some one please show me how to show that this equation is true?

cot^2x-cos^2x= cot^2x * cos^2x

Consider the $\displaystyle LHS$.

$\displaystyle \cot^2x - \cos^2 x = \frac {\cos^2 x}{\sin^2 x} - \cos^2 x$

$\displaystyle = \cos^2 x \left( \frac 1{\sin^2 x} - 1 \right)$

$\displaystyle = \cos^2 x (\csc^2 x - 1)$

$\displaystyle = \cos^2 x \cot^2 x$

Quote:

and i have one more question:
in solving:
4cos^2x-3=0
i get +- sqrt. 3/2
this means that there are 4 answers, but in the book, it gives me 2 :
pi/6+npi and 5pi/6+npi
why?

thanks
no, there are infinitely many answers! you were not given any bounds. the answers given are the general formulas for ALL solutions
• May 6th 2008, 04:15 PM
>_<SHY_GUY>_<
so that would include the solutions in quads 3 and 4?
what is LHS?
• May 6th 2008, 04:19 PM
Jhevon
Quote:

Originally Posted by >_<SHY_GUY>_<
so that would include the solutions in quads 3 and 4?

because of the +/-, you will have solutions in all quadrants

$\displaystyle \frac {\pi}6 + n \pi$ takes care of all solutions in quadrants 1 and 3, while $\displaystyle \frac {5 \pi}6 + n \pi$ takes care of all solutions in quadrants 2 and 4
• May 6th 2008, 04:21 PM
icemanfan
Quote:

Originally Posted by >_<SHY_GUY>_<
so that would include the solutions in quads 3 and 4?
what is LHS?

LHS = Left-hand side

And yes, there are solutions in the 3rd and 4th quadrants. Notice the forms of the answers are given as $\displaystyle x + \pi n$ and not $\displaystyle x + 2\pi n$.
• May 6th 2008, 04:23 PM
>_<SHY_GUY>_<
so that would explain that it would alternate to quad 1 to quad 3...solution wise?(Bow)

how do you get that equation to look like that?