# Help to solve an equation

Discussion in 'Math' started by woon_h88, Nov 16, 2015.

1. ### woon_h88 Thread Starter Active Member

Mar 25, 2009
46
2
Hi,

I been stuck in this equation : (5pi/6 - x)sin x = cos x + 0.886

The finally answer for x = pi/3.

Everyone have any idea to solve?

Thank you

RRITESH KAKKAR likes this.
2. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Have you started the preliminary steps? It would help if you showed what you have done.

3. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Analytically?

I doubt you can. But notice the 0.886 term. Doesn't that scream out to you? Hint: It's sqrt(3)/2.

4. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Oh, and pi/3 is not an exact answer, but it's close.

[Much later edit]: Actually, pi/3 IS an exact answer. This is because it forces the left hand side to be zero and, as a result, the fact that the angle is both outside and inside a transcendental function becomes immaterial at that point.

Last edited: Dec 2, 2015
5. ### woon_h88 Thread Starter Active Member

Mar 25, 2009
46
2
Hi,
I try like
(5pi/6 - x)sin x = cos x + 0.886
5pi/6 sin x - x sin x = cos x + 0.886
5pi/6 sin x = cos x - x sin x + 0.886

But seem too lost, so i try,
(5pi/6 - x)sin x = cos x + 0.886
(5pi/6 - x)sin x = sin ( x + pi/2) + 0.886
5pi/6 sin x - x sin x - sin ( x + pi/2) = 0.886
By using trigo formula, Im back to:
5pi/6 sin x - x sin x -cos x = 0.886

Needed some tips to save me out of this while loop.

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
You are not going to be able to solve it analytically. You will need to use numerical methods.

7. ### wayneh Expert

Sep 9, 2010
12,376
3,226
Ooh, are you sure of that? I'd be reluctant to declare that without a lot more work on it. (I haven't spent any time on it, so you may well be right.)

I would ask the TS: If this is homework. what topics have you been studying lately? For instance, if you've just been taught the law of cosines, it might be prudent to apply that here. I'm not saying this problem has anything to do with that law, just making the point that you may have recently been given the tools you need to solve this.

8. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I'm not absolutely sure, but I did play around with it some. The answer is definitely NOT exactly pi/3 (unless I really screwed up), so that makes me more confident that it is non-analytic.

9. ### J_Rod Member

Nov 4, 2014
109
6
The first step is to determine the interval you're interested in finding x such that both sides of the equation are equal, or x such that
$\left (\frac {5 \pi} {6} -x \right) sin(x) -cos(x) -0.886 = 0$
considering the solution set is infinite and aperiodic. Then you can graph this and guess where the zeroes are, and maybe apply the Bisection method or Newton's method for a good approximation to desired precision.

Note:
$\(1/2) sqrt {3} \approx 0.866$

10. ### wayneh Expert

Sep 9, 2010
12,376
3,226
Bingo. But you can see that the 5π/6 term is negligible at large x, where the function resembles: -x•sin(x)-cos(x)-0.866 = 0 and the zero crossings are pi apart.

11. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I think the zero crossings might approach being pi apart in the asymptotic limit, but they can't be exactly pi apart otherwise that multiplicative x term will trash things unless sin(x) happens to be zero. But we know that isn't the case because then cos(x) will either be 1 or -1.

As x grows, sin(x) has to get closer and closer to zero to keep in in check, but that means that x has to move a bit off of an increment of pi in order to let the cos(x) term eat up the new difference.

If we look at the original function, we see that it is an even function (since x·sin(x) is even and cos(x) is even), meaning that any solution for x>0, there is also a solution at -x.

However, unless otherwise stated, the solution sought is almost certainly the principal solution, which is usually the smallest solution greater than or equal to zero for a problem like this.

12. ### wayneh Expert

Sep 9, 2010
12,376
3,226
The answer does flip flop from just under an integral factor times pi to just over, getting to within 1% of pi by x=20.
My math teacher would have given a zero to anyone not noticing there was more than one solution.

13. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
In a math class, I definitely understand expecting the failure to notice multiple solutions not being accepted.

It all depends on context. In almost all situations (even in most math contexts), if asked what the square root of something is it is almost always expected that only the principal square root is being sought. The cue of whether the negative square root should be considered has to be taken from the context of the problem. This is reflected in such basic things as the quadratic formula. If the square root was always expected to be understood to include both positive and negative solutions, then there would be no need for the +/- symbol that seems to always be present before the radical.

14. ### woon_h88 Thread Starter Active Member

Mar 25, 2009
46
2
Thank for all the reply.. Its one of a question from power system analysis..I used iterations method as say by J_Rod by sub an pre-fault angle into the "x" until the value converge...Even though its might be long but its quite easy to remember for my exam..>.<

15. ### wayneh Expert

Sep 9, 2010
12,376
3,226
Without making a plot, I think that's a dangerous approach, IF you are expected to find the smallest positive solution as WBahn mentioned. See how easy it would be to choose the wrong zero. (Note that both of my charts plot the RHS of J_Rod's equation vs. x).

16. ### MrAl Distinguished Member

Jun 17, 2014
2,553
515
Hi,

It is interesting that if you make the constant 0.886 equal to sqrt(3)/2 and take the derivative and solve that for zero, you get a local min which is coincident with an exact solution which is at 5*pi/6, for what it is worth.
The other derivatives would help in setting up limits for finding other solutions.

17. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
That's actually a special case that someone (myself included) should have spotted quite some time ago. What happens if you drive the left hand side to zero identically? Just wasn't thinking along those lines when I first saw the problem and then got funneled down a particular approach once I had concluded that it was non-analytic.

18. ### MrAl Distinguished Member

Jun 17, 2014
2,553
515
Hi again,

If that is what i did a little while ago then you get 5*pi/6 again. That *is* a solution though.

But i may have found the analytical approach. It comes after converting the equation into a quadratic in cos(x), and then that can be solved for y (where y=cos(x)) and then the solution is x=acos(y).
It's a little involved so i will go over it again before posting. I got a number close to pi/3 but pi/3 isnt exact anyway.
Anyone else may want to try this too and see what they get.

I see what happened, i started from an assumed solution so that idea doesnt work.

So scratch that, but next idea would be to try to turn it into a conic in x and either cos(x) or sin(x), then rotate, then solve for x.
Note: doesnt work either.

Last edited: Dec 2, 2015
19. ### MrAl Distinguished Member

Jun 17, 2014
2,553
515
Hello again,

After trying a few methods like that, none of them work out probably because of that lone 'x' that multiples the sine function like x*sin(x). That's hard to deal with when there are other terms too. I dont know of a method that will transform this into a solvable equation, but there could be one somewhere that we are overlooking.
If it were not for that 'x' multiplier this would be more easily solvable. Without that 'x' it can be transformed into a quadratic as mentioned earlier.
BTW the attempt to transform into a conic resulted in a non conic with at least one offending term: x^2*y^2 and no way to reduce that to just x*y.

20. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
I'm pretty sure that the equation is non-analytic, precisely because of that x·sin(x) term. I'm also pretty sure that the fact that the equation was carefully crafted such that one of the solutions effectively removes the offending term does not change the fact that the equation as a whole is non-analytic.