- #1

- 1

- 0

18sinxcosx=1-6sinx

Would there be an easy way other than trial and error?

Thanks

- Thread starter O garcia
- Start date

- #1

- 1

- 0

18sinxcosx=1-6sinx

Would there be an easy way other than trial and error?

Thanks

- #2

dynamicsolo

Homework Helper

- 1,648

- 4

Rather than just try solutions one by one, you could plot f(x) = 1 - 6 sin x - 9 sin 2x and locate the zeroes of the curve. There are four solutions in the interval [0, 2·pi]. The good news is that the function has a period of 2·(pi), so once you find those four, you can describe all the rest.

- #3

- 210

- 0

scrap this

Last edited:

- #4

- 350

- 1

- #5

Dick

Science Advisor

Homework Helper

- 26,260

- 619

- #6

- 238

- 0

By trigonometric sine rule, sin(a)cos(b)=(1/2)[sin(a+b)+sin(a-b)]

Edit: sorry, I'll use a and b to avoid confusion. Since a=b=x, then

Simplify this equation, we get (1/2)[sin(2x)+sin(x-x)]=sin(2x)/2 (*)

Plug (*) into the LHS, then we get 9sin(x)=1-6sin(x)

Solve, for x. It should be easy from here. =]

Edit: sorry, I'll use a and b to avoid confusion. Since a=b=x, then

Simplify this equation, we get (1/2)[sin(2x)+sin(x-x)]=sin(2x)/2 (*)

Plug (*) into the LHS, then we get 9sin(x)=1-6sin(x)

Solve, for x. It should be easy from here. =]

Last edited:

- #7

- 210

- 0

Hmm..I posted something similar to that but deleted it since I set the coefficient of -a*sin(x) to 1 instead of 6.

One thing to note though, if you fling this into a CAS...you get a very, ahem... 'clever looking' solution. (ie; ridiculously messy!!!) :tongue2:

- #8

dynamicsolo

Homework Helper

- 1,648

- 4

You dropped a '2' in that last equation: that should beBy trigonometric sine rule, sin(x)cos(x)=(1/2)[sin(x+x)+sin(a-x)]

Simplify this equation, we get (1/2)[sin(2x)+sin(x-x)]=sin(2x)/2 (*)

Plug (*) into the LHS, then we get 9sin(x)=1-6sin(x)

Solve, for x. It should be easy from here. =]

9 sin(

which is what I gave back in post #2. Unfortunately, there isn't any

Alas, I looked at that possibility and it doesn't work for this equation. An approach of that sort will work for certain special cases of the numerical coefficients -- but this ain't one o'them...

Last edited:

- #9

- 238

- 0

Ah, sorry. I didn't see that.You dropped a '2' in that last equation: that should be

9 sin(2x) = 1 - 6 sin(x) ,

which is what I gave back in post #2. Unfortunately, there isn't anyniceway to solve that...

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 878

- Replies
- 6

- Views
- 979

- Replies
- 4

- Views
- 2K

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 784

- Replies
- 3

- Views
- 809