[SOLVED] Trig

**>_<SHY_GUY>_<** · May 6th 2008, 04:05 PM

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

**Jhevon** · May 6th 2008, 04:13 PM

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

**>_<SHY_GUY>_<** · May 6th 2008, 04:15 PM

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

**Jhevon** · May 6th 2008, 04:19 PM

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

**icemanfan** · May 6th 2008, 04:21 PM

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 $x + 2\pi n$ .

**>_<SHY_GUY>_<** · May 6th 2008, 04:23 PM

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?

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