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# C4 Solve: sin4theta = cos 2theta watch

1. solve the following equation for -Pi<0<Pi
sin4theta = cos 2theta

I tried sin(2theta+2theta) but still

Thx
2. I reckon this'll work; I'll be using x instead of theta as it's more concise.

sin(4x)=cos(2x)
expand LHS using sin double angle formula
2sin(2x)cos(2x)=cos(2x)
divide both sides by cos(2x)
2sin(2x)=1
divide by two
sin(2x)=1/2

can you take it from here? (hint: I think there are four solutions in the range...)

Note: I think this is right, but haven't done this for a while now...
If it is wrong the error will be from dividing by cos(2x) - do reply and let me know if this is or isn't valid.

Hope this helps!
3. (Original post by alan.accordion)
I reckon this'll work; I'll be using x instead of theta as it's more concise.

sin(4x)=cos(2x)
expand LHS using sin double angle formula
2sin(2x)cos(2x)=cos(2x)
divide both sides by cos(2x)
2sin(2x)=1
divide by two
sin(2x)=1/2

can you take it from here? (hint: I think there are four solutions in the range...)

Note: I think this is right, but haven't done this for a while now...
You can't (or shouldn't) divide by cos2x here. It's part of the answer you're looking for, so by dividing through by it, you're elminating (potentially) 2 solutions.

I would keep extrapolating by using the double angle formulae. It will reduce, eventually, to an equation containing only one circle function.
4. (Original post by alan.accordion)
I reckon this'll work; I'll be using x instead of theta as it's more concise.

sin(4x)=cos(2x)
expand LHS using sin double angle formula
2sin(2x)cos(2x)=cos(2x)
Up to here is right, but remember cos(2x) could be equal to 0 so you can't divide through by it.
5. (Original post by mufas)
solve the following equation for -Pi<0<Pi
sin4theta = cos 2theta

I tried sin(2theta+2theta) but still

Thx
did this in 30 secs :P. Not the solution but the method. Sin4A= 2sin(2A). Cos (2A)

now sin 4A = cos 2A, thus: 2sin 2A.cos2A-Cos2A = 0

Factorising: Cos 2A*(2Sin2A -1) = 0.

You solve this between the limits.
6. (Original post by blue_shift86)
did this in 30 secs :P. Not the solution but the method. Sin4A= 2sin(2A). Cos (2A)

now sin 4A = cos 2A, thus: 2sin 2A.cos2A-Cos2A = 0

Factorising: Cos 2A*(2Sin2A -1) = 0.

You solve this between the limits.
huh ? Didn't get it
7. (Original post by mufas)
huh ? Didn't get it
Essentially, he's saying you can take the Cos2x over to the other side, expand Sin4x (Sin(2x+2x)using the double angle identity, and take out Cos2x as a factor.

It's a legitimate way to do it. You just have to remember that each trig function you're doing is in the interval 0<x<4pi, so you're going to have a fair few solutions, (i.e, go through 2 periods of the respective trig graphs, or around the unit circle twice).
8. Hint: note that and recall that if then or .
9. (Original post by mufas)
huh ? Didn't get it
Apologies, let me try again:

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

Sin (A+B) = SinA. CosB + SinB. Cos A. [Sin two angle identity)
If you let A and B both be 2x, then you get:

LHS: Sin (2x+2x) = Sin(2x).Cos(2x) + Sin(2x).Cos(2x) = 2Sin(2x).Cos(2x)
RHS = cos (2x)

Now LHS=RHS

2Sin(2x). Cos(2x) = Cos (2x) [take cos 2x to other side)
2Sin(2x).Cos(2x) - Cos (2x) = 0
Now factorise out Cos (2x)

So: Cos(2x).[2Sin(2x) - 1] = 0

Now you just solve it normally between -pi and +pi.

I hope this helps. Sorry for being a bit late. I've been gardening for the last few hours
10. (Original post by Piecewise)
Hint: note that and recall that if then or .
Give over! If the opening poster couldn't follow the solutions given earlier in the thread, they'll have no chance with that.
11. (Original post by blue_shift86)
Apologies, let me try again:

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

Sin (A+B) = SinA. CosB + SinB. Cos A. [Sin two angle identity)
If you let A and B both be 2x, then you get:

LHS: Sin (2x+2x) = Sin(2x).Cos(2x) + Sin(2x).Cos(2x) = 2Sin(2x).Cos(2x)
RHS = cos (2x)

Now LHS=RHS

2Sin(2x). Cos(2x) = Cos (2x) [take cos 2x to other side)
2Sin(2x).Cos(2x) - Cos (2x) = 0
Now factorise out Cos (2x)

So: Cos(2x).[2Sin(2x) - 1] = 0

Now you just solve it normally between -pi and +pi.

I hope this helps. Sorry for being a bit late. I've been gardening for the last few hours
Thank you soo much!!!
12. (Original post by alan.accordion)
I reckon this'll work; I'll be using x instead of theta as it's more concise.

sin(4x)=cos(2x)
expand LHS using sin double angle formula
2sin(2x)cos(2x)=cos(2x)
divide both sides by cos(2x)
2sin(2x)=1
divide by two
sin(2x)=1/2

can you take it from here? (hint: I think there are four solutions in the range...)

Note: I think this is right, but haven't done this for a while now...
If it is wrong the error will be from dividing by cos(2x) - do reply and let me know if this is or isn't valid.

Hope this helps!
thank you There are 8 answers between -pi<0<pi
13. (Original post by mufas)
thank you There are 8 answers between -pi<0<pi
no problem! that's why i hate questions like this, lol! Too many sinver cos and sin crap to do! :P and too much +ing and -ing :P
14. Hmm, instead of that division of cos(2x) earlier, you could just factorise it and go from there if I'm not mistaken.
15. (Original post by Piecewise)
Hint: note that and recall that if then or .
yer yer, quit showing off now douche bag.
16. Jeez, was my hint really that hard?

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