Question

How should someone start trying to find the solutions of sin 2 x - sin 4 x?

I have some ideas on how to start, but I'm not sure what would be the right way to approach this. Would you extract sin 2 x to get sin 2 x (sin x - sin 2 x) and go from there, or would you simplify sin 2 x to 2 sin x cos x to get (2 sin x cos x) - sin 4 x? And could you explain why that's so?

Answer 1

HI!!

Answer 2

\[\sin^2(x)-\sin^4(x)\]?

Answer 3

sin(4x) = sin(2*2x) = 2 sin(2x) cos (2x)

Answer 4

not really clear, and in any case it is not an equation, so there is nothing to "solve"

Answer 5

maybe \[\sin^2(x)-\sin^4(x)=0\]? not clear
you got a screen shot?

Answer 6

@misty1212 Nope. It'd be \[\sin 2 x - \sin 4 x\]

@satellite73 what do you mean it can't be solved? It does have multiple solutions, so there's no one solution if that's what you meant. I'm working within the intervals [0, 2pi] if that helps.

Answer 7

you can solve an equation like \[2x-1=0\] you cannot solve \[2x-1\]

Answer 8

@satellite73 oh, i get what you mean. yeah. in my case sin 2 x - sin 4 x does equate to 0.

Answer 9

The problem is, your mathematical statement is NOT an equation and thus cannot be solved. If you append "=0" to your statement, then it becomes solvable.

Answer 10

Putting that "=0" in there is NOT optional. It's mandatory.

Answer 11

@mathmale I see. I should've worded that properly, my bad.

Answer 12

Glad to hear this.

Answer 13

Note that sin 4x can be re-written as sin (2x + 2x). Look up the summation formula for the sine function: sin 2x = ???

Answer 14

Then apply that formula to sin 4x.

Answer 15

@mathmale Summation formula? I don't think my school mentioned anything like that! Well, I can work with this. This makes my equation \[\sin 2 x - \sin (2x + 2x)\], right? So from here, do I redistribute the sin 2 x to get something like:\[\sin 2 x - \sin 2 x + \sin 2 x\]?

Answer 16

Look up the formula for sin (a + b). That's the "summation formula" I'm referring to.

Answer 17

Wait a minute, it'd be \[\sin 2 x - \sin 2 x - \sin 2 x\] because of the -1

Answer 18

Not so. Please look up "trig formulas" and don't risk getting led astray by guessing.

Answer 19

OK, I see. Would you use a sum and difference formula here?

Answer 20

sin (a+b) = sin a cos b + cos a sin b. You'll need this formula in the future; may as well make a note of it and review it every now and then. Yes, you could call this the "sum and difference formulas" for the sine.

Answer 21

Apply this formula to expand sin (4x) = sin (2x + 2x).

Answer 22

Ah, so they did teach this to me! we simply called a and b x and y. OK, This gives me:
\[\sin 2 x - \sin 2 x \cos 2 x + \cos 2 x \sin 2 x\]

Answer 23

Have to put parentheses around the last 2 terms:\[\sin 2 x - [\sin 2 x \cos 2 x + \cos 2 x \sin 2 x]\]

Answer 24

Note that sin 2x is now a factor of all three terms. What next?

Answer 25

Factor out sin 2 x?

Answer 26

Be careful with your signs. Those parentheses do make a difference.

Yes. Factor sin 2x out. and simplify the results.

Answer 27

I know this sounds silly, but can I ask how you'd factor out this kind of equation? I haven't tried to factor out something like this before, I believe.

Would it go something like this?:

sin 2 x (-[cos 2 x + cos 2 x])

This is just my guess, but of course I can't be sure that's right.

Answer 28

sin 2x divided by sin 2x is 1, right? Where's your '1'? What is -(cos 2x + cos 2x)?

Answer 29

Oh, OK. So should this be

\[\sin 2 x (1 - (1 \cos 2 x + 1 \cos 2 x)\] ?

Answer 30

Yes, that's partially right. But what is 1cos 2x + 1cos 2x?

Answer 31

2 cos 4 x?

Answer 32

What is 1 egg + 1 egg?
Same principle applies to 1cos 2x + 1cos 2x.

Answer 33

So, 2 cos 2 x.

Answer 34

And the cos 2 x doesn't change when you add 2 of the same kind together? Of course.

Answer 35

Much better. Then you have sin 2x (1-2cos 2x).
You must set this =0 and then solve for x. There will be more than one solution.

Answer 36

Yup, this was what I needed. Thanks!

Answer 37

My pleasure.
Good luck.