How would i compute x?

[O garcia]

Hey can anyone help me? I'm trying to compute x in this equation:
18sinxcosx=1-6sinx

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

 

[dynamicsolo]

Since this could be rewritten as 1 - 6 sin x - 9 sin 2x = 0 , I don't think there's a tidy way to solve for x . The solutions are periodic (so there are an infinite number of them), but don't appear to take on obvious values.

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.

 

[GregA]

scrap this

 

[DavidWhitbeck]

If you square both sides and plug in $\cos^2 x = 1 - \sin^2 x$ then you will obtain a quadratic in sin, which you can solve using the quadratic formula.

 

[Dick]

Uh, you get a QUADRIC equation, which you can solve by the QUADRIC formula. Looks pretty nasty to me.

 

[konthelion]

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. =]

 

[GregA]

If you square both sides and plug in $\cos^2 x = 1 - \sin^2 x$ then you will obtain a quadratic in sin, which you can solve using the quadratic formula.

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:

 

[dynamicsolo]

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

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

If you square both sides and plug in $\cos^2 x = 1 - \sin^2 x$ then you will obtain a quadratic in sin, which you can solve using the quadratic formula.

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...

 

[konthelion]

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 any nice way to solve that...

Ah, sorry. I didn't see that.