Question

Can someone help me understand this?
http://puu.sh/3dy2d.png
http://puu.sh/3dy0I.png

options:
http://puu.sh/3dy2U.png

Answer 1

Yes! Simplify the expression first. Now what?

Answer 2

I'm totally lost.

Answer 3

Ok my bad, let's back up :)

Answer 4

Hi Please find attached

Answer 5

\[\large \frac{\sin^2\theta}{\cos^2\theta}=-\frac{3}{2}\frac{1}{\cos \theta}\]

Multiplying both sides by cos^2 gives us,\[\large \sin^2\theta=-\frac{3}{2}\cos \theta\]

Understand how the cosines cancelled on the right?

Answer 6

@zepdrix u should never cancel the common terms. You will lost a solution by doing so

Answer 7

@Anu2401 thank you so much! I'm going to let @zepdrix keep explaining though.

Yes I see how the cosines canceled on the left, and how we're left with cos instead of 1/cos.

Answer 8

Hmm I don't see what you're saying anu, this method will lead us to the exact same quadratic that you have.

Answer 9

@Anu2401 I've got one more I need help with besides this one, it's the same type of question though. Can you help with that too?

Answer 10

I"ll try :)

Answer 11

Remember this identity? it's one of the first ones you learn in trig :)

\(\large \color{royalblue}{\sin^2\theta+\cos^2\theta=1}\)
Subtracting cos^2 from each side,
\(\large \color{royalblue}{\sin^2\theta=1-\cos^2\theta}\)

We'll use this identity from where we're at,\[\large \color{royalblue}{\sin^2\theta}=-\frac{3}{2}\cos \theta \qquad \rightarrow \qquad \color{royalblue}{1-\cos^2\theta}=-\frac{3}{2}\cos \theta\]

Answer 12

@zepdrix : u r right . Yes in this case it doesn't matters but in quiet a few cases you will loose some general solution :)

Answer 13

@zepdrix Makes sense. Since sin^2(theta and 1-cos^2(theta) are equal, the substitution makes sense.

Answer 14

@Anu2401 http://puu.sh/3dzlX.png

thank you so much! you're a life saver.

Answer 15

From here, move some stuff around,\[\large \cos^2\theta-\frac{3}{2}\cos \theta-1=0\]Multiply both sides by 2 to get rid of that fraction,\[\large 2\cos^2\theta-3\cos \theta-2=0\]

And from here what we have is a quadratic! :O

Answer 16

It might look a little easier to handle if you make a substitution (like Anu did).


Let \(\large \cos\theta=x\), giving us,\[\large 2x^2-3x-2=0\]Remember how to handle something like this? :)

Answer 17

yep. do to one side what you'd do to the other

2x^2 - 3x - 2 + 2 = 0 + 2

2x^2 - 3x = 2
but how do you find x?

Answer 18

Woops, no no. Let's not move stuff to the right.

We have a quadratic.
So we'll either try to factor it, or if that's not possible, we'll throw it into the Quadratic Formula.

Answer 19

I've never seen my book mention a quadratic formula. O_O

Answer 20

Remember the quadratic formula from algebra?\[\large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 21

Luckily we won't need to use it, this one factors nicely.

Answer 22

Hmmm I'm always at a loss as to how to explain factoring... :\

Answer 23

I can show you the factors that it breaks into, but I'm having trouble finding a good way to explain it.

Answer 24

\[\large 2x^2-3x-2=0\]
Factors into,\[\large (2x+1)(x-2)=0\]

Answer 25

makes sense, because 2x^2 - 3 would equal out to 2x + 1, right? and x-2 is left the same. so it all comes down to 3x - 1?

Answer 26

@zepdrix is right just remember the formula. Its hard to find the roots sometimes .

Although in this case

2x^2 - 3x -2=0

Step1: Observe that what is the product of coifficient of x^2 & constant term (Both considered with their sign)
Here it is -4

Step2: Now find the factors of -4 such that when u multiply them u get -4 & when u add them u get -3

Then proceed as shown in the page scan0007.jpg.

For the next one i think this should suffice (PFA)

Answer 27

@zepdrix @Anu2401 thank you both so much!