Hello!

This is the equation Im stuck with:

Im not sure where to start, because I cant take out any of the x values,

should i let 0.5x be a completely new vale like 'y' and but then i will still be stuck

Thank you!

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- Apr 23rd 2010, 12:32 AMappleseedTricky sin equation
Hello!

This is the equation Im stuck with:

Im not sure where to start, because I cant take out any of the x values,

should i let 0.5x be a completely new vale like 'y' and but then i will still be stuck

Thank you! - Apr 23rd 2010, 12:51 AMsa-ri-ga-ma
- Apr 23rd 2010, 01:01 AMappleseed

Hello!

Thank you so much for the reply!!!

Yes i am asked to find the value of x and the answer is required to 3 s.f.

It might be better if i give you the context of the equation.

''''''''''''There are two equations g(x) and f(x), There are two values of x for which the gradient of f is equal to the gradient

of g. Find both these values of x. '''''''''"

So i've already found the derivative of the two equations and i've checked that its correct, and then i equated the two derivatives, but i dont know how to find x.

I was thinking that maybe it could be done by trig identities, lol but i dont know how though

anyways Thank you! !!(Rofl) - Apr 23rd 2010, 11:35 AMGrandad
Hello appleseed

There are infinitely many solutions to the equation

Have a look at the diagram I've attached, where I've plotted the graph of each side separately.

You won't be able to solve an equation like this to get exact answers. You'll have to use a numerical method - e.g. Newton-Raphson. Do you know how?

Grandad - Apr 24th 2010, 10:14 PMappleseed
- Apr 24th 2010, 10:44 PMGrandadNewton-Raphson method
Hello appleseedIf you've not had any practice with this method, this is not the easiest example to start with.

If you want to get a solution to the equation

The next approximation, , uses as the starting value:

(to 2 d.p.)

The next positive solution is around . To 4 d.p., the solution converges very quickly to .

Grandad