(05-26-2016, 07:47 PM)Psycho_Coder Wrote: @Jolly

I will give you some hint.

Code:

sin \left(x \right ) = \cos \left(2x \right ), [0, 2\pi]
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Using \hspace{2 mm} the \hspace{2 mm} identity,\hspace{2 mm} \cos \left(2x \right) = 1 - 2\sin^{2} \left(x \right)
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Hence, \hspace{2 mm} using \hspace{2 mm} the \hspace{2 mm} above \hspace{2 mm} in \hspace{2 mm} the \hspace{2 mm} given \hspace{2 mm} equation.
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sin \left(x \right ) = 1 - 2\sin^{2} \left(x \right)
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\Rightarrow 2\sin^{2} \left(x \right) + sin \left(x \right ) - 1 = 0
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Solving \hspace{2 mm} the \hspace{2 mm} above \hspace{2 mm} quadratic \hspace{2 mm} equation \hspace{2 mm} we \hspace{2 mm} get \hspace{2 mm},
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(2\sin \left(x \right) - 1) (\sin \left(x \right) + 1) = 0

I haven't solved fully but I am giving you a basic idea. If you solve the final equation you will get three solutions for x. Try and see for yourself. Put the above latex code here and convert to see the output.

Sample:

(05-24-2016, 10:02 PM)Eclipse Wrote: Sorry, was away.

For your first question, assuming the area under the graph between x=b and x=m is equal to the area under the graph between x=m and x=a, and assuming you know the values of a and b, you could actually find the integral of y=f(x), and then substitute in the values and equate them.

i.e.
let g(x) be the function of the integral of f(x)
thus g(b) - g(m) = g(m) - g(a)

Then rearrange and solve for m. If you come out with a polynomial, you can find the discriminant and check if it's more than or equal to 0. If not, there's no point solving for m since there's no (real) solutions.

As for your second question, I actually don't know what you mean. Maybe clarify it a little?

(06-04-2016, 03:34 PM)Rick Wrote:

(05-24-2016, 10:02 PM)Eclipse Wrote: Sorry, was away.

For your first question, assuming the area under the graph between x=b and x=m is equal to the area under the graph between x=m and x=a, and assuming you know the values of a and b, you could actually find the integral of y=f(x), and then substitute in the values and equate them.

i.e.
let g(x) be the function of the integral of f(x)
thus g(b) - g(m) = g(m) - g(a)

Then rearrange and solve for m. If you come out with a polynomial, you can find the discriminant and check if it's more than or equal to 0. If not, there's no point solving for m since there's no (real) solutions.

As for your second question, I actually don't know what you mean. Maybe clarify it a little?

God I looked at this thread and I didn't have a clue as to what I was talking about, my notation is just shocking.

I found this which was essentially what I was looking for, I could barely understand what I was talking about in that post don't worry.

(06-10-2016, 10:41 PM)Eclipse Wrote:

(06-04-2016, 03:34 PM)Rick Wrote:

(05-24-2016, 10:02 PM)Eclipse Wrote: Sorry, was away.

For your first question, assuming the area under the graph between x=b and x=m is equal to the area under the graph between x=m and x=a, and assuming you know the values of a and b, you could actually find the integral of y=f(x), and then substitute in the values and equate them.

i.e.
let g(x) be the function of the integral of f(x)
thus g(b) - g(m) = g(m) - g(a)

Then rearrange and solve for m. If you come out with a polynomial, you can find the discriminant and check if it's more than or equal to 0. If not, there's no point solving for m since there's no (real) solutions.

As for your second question, I actually don't know what you mean. Maybe clarify it a little?

God I looked at this thread and I didn't have a clue as to what I was talking about, my notation is just shocking.

I found this which was essentially what I was looking for, I could barely understand what I was talking about in that post don't worry.

I'm actually interested to find out if my method would work.