Question

Solving trigonometric equations (Precalc)

Answer 1

is there a certain equation that has been bugging you?

Answer 2

Well, starting with 15, I understand that I have to factor it, but I'm not sure how. I tried to factor out a sin but it wasn't going how I wanted it to

Answer 3

Well if you want to post the equation, I think I will able to help more. Otherwise I don't have enough information.

Answer 4

I posted the photo, here I can post it again.

Answer 5

That is the first attachment on this thread...But anyways your picture is do dark on top...I will try to make it out:
\[2 \sin^2(x)-5 \sin(x)-3=0\]
is that right?

Answer 6

if so do you know how to factor
\[2u^2-5u-3\]

Answer 7

Yes, and so I factor out a u in this case

Answer 8

well you can't there isn't a u in each term

Answer 9

if you don't know how to factor quadratics
you can always use the quadratic equation to solve
2u^2-5u-3=0

Answer 10

I mean formula

Answer 11

\[au^2+bu+c=0 \\ u=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 12

So you can factor out a 2?

Answer 13

Have you ever solved quadratic equations using the quadratic formula ? you don't have to factored anything
you just have to get in the form au^2+bu+c=0 which you already have

Answer 14

But I feel like that's extra work, is it easier to solve it that way? And yes I have

Answer 15

well you don't know how to factor 2u^2-5u-3
so I suggest using the quadratic formula to solve 2u^2-5u-3=0


unless you want me to teach you how to factor quadratic expressions?

Answer 16

au^2+bu+c can be factored over the integers
if you find two numbers that multiplied to be a*c and add up to be b
you have 2u^2-5u-3
a*c=2*(-3)=-6
are there two factors of -6 that have product -6 and sum -5?

Answer 17

That would be preferable but if you don't have the time then that's okay

Answer 18

-1 and 6

Answer 19

Ah I see, okay thank you. I actually know how to so this I was just mislead.

Answer 20

well I think you mean 1 and -6

Answer 21

replace -5u with -6u+1u

Answer 22

2u^2-6u+1u-3
then factor by grouping