Question

I need help with this.

Find ask real number solutions in the interval [0,2pi) that satisfy each equation. Round approximate answers to the nearest tenth.

2cos^2(2x)-8sin^2(x)cos^2(x)=-1

I don't see any identity substitutions or a way this factors. Hopefully I'm overlooking something.

Answer 1

I'm not sure what level of trig is required here but are you familiar with the identity
cos2x = cos^2(x) - sin^2(x)?
Knowing that, you can do some dirty maths with 2cos^2(2x), which is really just 2 times cos(2x) times cos(2x).

Answer 2

I'm familiar with it. I guess I hadn't noticed that.

Answer 3

Are you able to go on from there?

Answer 4

Maybe, I'm going to try.

Answer 5

I was unable to solve it.

Answer 6

What did you get up to? It's quite a hard question really, either that or I did it a super weird way, but after expanding (cos(2x))^2 and putting it back into the equation, I was able to simplify things to get
cos^2(x) sin^2(x) = 1/4, and then I used the trig identity sin2x = 2sinxcosx to simplify things even further

Answer 7

I think I went a different way.
so, you had
2[cos^2(x)-sin^2(x)][cos^2(x)-sin^2(x)]-8sin^2(x)cos^(x)=-1

Did you factor out a [cos^2(x)-sin^2(x)]?

Answer 8

Not quite. I just multiplied them together, using the fact that (a-b)^2 = a^2 - 2ab + b^2

Answer 9

It looks like a disgusting way of doing it (and it might not be the fastest way, idk) but we would end up with cos^4(x) + sin^4(x) which is actually just equal to 1

Answer 10

I mean, cos^4(x) + sin^4(x) in addition to a -2cos^2(x)sin^2(x)

Answer 11

so

Answer 12

so
2cos^4(x)-2cos^2(x)sin^2(x)+sin^4(x)-8sin^2(x)cos^2(x)=-1

2cos^4(x)+sin^4(x)-2cos^2(x)sin^2(x)-8sin^2(x)cos^2(x)=-1

I still don't see it.

Answer 13

You made a bit of an error. The 2 outside the cos^2(2x) applies to its entire expansion. So it would actually be:
2(cos^4(x) - 2cos^2(x)sin^2(x)+ sin^4(x)) -8sin^2(x)cos^2(x)=-1
2cos^4(x) - 4cos^2(x)sin^2(x)+ 2sin^4(x) -8sin^2(x)cos^2(x)=-1

Now we know that sin^2(x) + cos^2(x) = 1. The thing is, sin^4(x) + cos^4(x) = 1 too (you can go about proving that to yourself later if you like).

Answer 14

You applied that identity after or before multiplying the two back in??

Answer 15

Well, before really, but I was just showing you what it should look like if you multiplied through by 2.
In my own working I had it as
2(cos^4(x) - 2cos^2(x)sin^2(x)+ sin^4(x)) -8sin^2(x)cos^2(x)=-1
2(1 - 2cos^2(x)sin^2(x)) -8sin^2(x)cos^2(x)=-1

Answer 16

Actually to be honest I'm beginning to doubt that sin^4(x) + cos^4(x) is really equal to 1...
sorry that's my bad :(

Answer 17

1 times 1 is 1. I'm willing to use it for now. I can email a solution to my teacher tomorrow so she can check it.

Answer 18

Sorry but it's not actually equal to 1, I was getting mixed up with something else.

I don't really have time to finish helping with this question, but if I was you I would write down every trig identity you know and try and see how you can apply a combination of them in the question to simplify it down into a situation where you only have 1 type of trig function.
Then, you might be able to use a substitution like u = cos^2(x) and factorise your new u equation.

Sorry about that.

Answer 19

Thanks for trying.