Help with trigonomic equation...

i am supposed to find all solutions in [0,2pi). the original equation is

4sin^2x - 3sinx - 2 = 0

i started off using the quadratic formula and got

3 +/- sqrt|-3^2 - 4*4*-2 / 2*4

after simplifying all these through i was left with 3 +/- sqrt41/8

this is where i am confused and really have no idea where to continue, there really isn't a problem to explain this in the book so any help would be great.

Re: Help with trigonomic equation...

Put these in your calculator

$\displaystyle \displaystyle \sin^{-1}\left( \frac{3+ \sqrt{41}}{8}\right)$

and

$\displaystyle \displaystyle \sin^{-1}\left( \frac{3- \sqrt{41}}{8}\right)$

What do you get?

Re: Help with trigonomic equation...

do you mind explaining how to do that?

Re: Help with trigonomic equation...

I accidently edited post #2

Re: Help with trigonomic equation...

when i do http://latex.codecogs.com/png.latex?...%7D%5Cright%29 i get domain error

when i do the other one i get -25.1754 rounded

Re: Help with trigonomic equation...

Quote:

Originally Posted by

**lolcheelol**

do you understand how the inverse trig functions work and how to interpret what the calculator is telling you?

Pauls Online Notes : Calculus I - Review : Trig Equations with Calculators, Part I

Re: Help with trigonomic equation...

Quote:

Originally Posted by

**skeeter**

yes i do, but usually i get stuff like like sin^-1(sqrt3/2) and ill end up with pi/3. so im not exactly sure how this answer is supposed to be written out. i can't find anything like it in my notes and in the book :/

Re: Help with trigonomic equation...

Quote:

Originally Posted by

**lolcheelol** yes i do, but usually i get stuff like like sin^-1(sqrt3/2) and ill end up with pi/3. so im not exactly sure how this answer is supposed to be written out. i can't find anything like it in my notes and in the book :/

you're used to solving equations whose solutions are unit circle values ... this equation requires the use of a calculator, hence the reason for the link. have a look.

Re: Help with trigonomic equation...

awesome, thanks a lot for the help. i had done a little bit of that in my notes, but nothing that explained. GREATLY appreciated!