- #1

- 1

- 0

18sinxcosx=1-6sinx

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

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- 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

- 26,263

- 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

By 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. =]

You dropped a '2' in that last equation: that should be

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...

Share: