Question

Find all solutions of the equation 3csc^2x - 4 = 0

Answer 1

\[3csc^{2}x = 4\]
\[ csc^2(x) = \frac{4}{3}\]
\[\frac{1}{sin^{2}x} = \frac{4}{3}\]
\[ 3 = 4sin^{2}x => 4sin^{2}x = 3 => sin^{2}x = \frac{3}{4}\]

Answer 2

You able to do it now?

Answer 3

Um. So would all of the solutions be: 60 + 360n ?

Answer 4

Sure... are you sure it isn't +180n instead?

Answer 5

Wait why would it be +180n?

Answer 6

Because..., for instance, 60 + 180
= 240

Try sin(240)

Answer 7

Right, I know it's going to give you something negative, but remember, we're looking for the solutions to
\[\large \sin^2(x)=\frac{3}{4}\]

and not just
\[\large \sin(x) = \sqrt{\frac34} =\frac{\sqrt3}{2}\]

Answer 8

So actually, a more specific way to write it would be finding the solutions to
\[\Large \sin(x) = \pm \frac{\sqrt3}{2}\]

Answer 9

would the answer for negative \[\sqrt{3}/2\] be the same as its positive?

Answer 10

No... but we're looking for the solutions to the equation
\[\large \sin^2(x) = \frac34\]
right?

Which means that if we take the square root of both sides...
\[\Large \sin(x) = \pm\sqrt{\frac34}\]we have to consider both the positive and negative square roots, since we are going to square it...

Answer 11

you can also look at it this way...
\[\Large \sin^2(x) - \frac34 = 0\\\\\\\\\Large let \ u=\sin(x) \]

\[\Large u^2 - \frac34 = 0\]
\[\Large \left(u-\sqrt \frac34\right)\left(u+\sqrt \frac34\right)=0\]

Answer 12

Oh ok. I still don't really get this concept though. For example, if I have.... 4 cos^2 x - 1 = 0

How would I find all of the solutions?

Answer 13

I'm sure it would be a very similar problem... if it makes things easier, you can always
let \(\Large u = \cos(x)\)

So it becomes
\[\Large 4u^2 -1 = 0\]
and then solve for u first...

Answer 14

What would you do after you solve for u and get +/- 1/2 ? How would you go about in finding all of the solutions?

Answer 15

Well, replace u again with cos(x)
\[\huge \cos(x) = \pm \frac12\]
and solve for all values of x whose cosines are either 1/2 or -1/2

Answer 16

Would that be 60 + 300n?

Answer 17

300n? You mean 360n?

Answer 18

Wait why would it be 360n?

Answer 19

Why would it be 300n?
360n is more understandable, as rotating 360 degrees essentially brings it back to the same angle...

300n?
at n = 1, it already doesn't hold...
cos(60 + 300) = cos(360) = 1

Answer 20

But actually, much like earlier, it's 60 + 180n

Answer 21

and also 120 + 180n

Answer 22

Oh ok. I see. Thank you!!!!

Answer 23

No problem :)