1. ## [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

2. 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 $LHS$.

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

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

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

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

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

3. so that would include the solutions in quads 3 and 4?
what is LHS?

4. 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

$\frac {\pi}6 + n \pi$ takes care of all solutions in quadrants 1 and 3, while $\frac {5 \pi}6 + n \pi$ takes care of all solutions in quadrants 2 and 4

5. 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 $x + \pi n$ and not $x + 2\pi n$.

6. so that would explain that it would alternate to quad 1 to quad 3...solution wise?

how do you get that equation to look like that?