[Psycho_Coder]

@Jolly

I will give you some hint.

Code:

sin \left(x \right ) = \cos \left(2x \right ), [0, 2\pi]
\\
\\
Using \hspace{2 mm} the \hspace{2 mm} identity,\hspace{2 mm} \cos \left(2x \right) = 1 - 2\sin^{2} \left(x \right)
\\
\\
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.
\\
\\
sin \left(x \right ) = 1 - 2\sin^{2} \left(x \right)
\\
\Rightarrow 2\sin^{2} \left(x \right) + sin \left(x \right ) - 1 = 0
\\
\\
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},
\\
\\
(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:

[Eclipse]

@Jolly

I will give you some hint.

Code:

sin \left(x \right ) = \cos \left(2x \right ), [0, 2\pi]
\\
\\
Using \hspace{2 mm} the \hspace{2 mm} identity,\hspace{2 mm} \cos \left(2x \right) = 1 - 2\sin^{2} \left(x \right)
\\
\\
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.
\\
\\
sin \left(x \right ) = 1 - 2\sin^{2} \left(x \right)
\\
\Rightarrow 2\sin^{2} \left(x \right) + sin \left(x \right ) - 1 = 0
\\
\\
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},
\\
\\
(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:

You could also just

let y = sin(x)
2y^2 + y - 1 = 0
So y = 1/2 and y = -1
So sin(x) = 1/2 and sin(x) = -1

And then solve from there using arcsin and a cast diagram/sin graph.

[Aeolian]

Why couldn't this thread be alive seven years ago when I was studying.

Netherless impressive thread Eclipse, I'll let you know when I need my government funds calculated.

[Psycho_Coder]

@Eclipse I expected the user to be able to do that much rather than simplifying it to the very basic level.

[Rick]

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.

[Eclipse]

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.

[Rick]

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.

I tried it with a few functions and it worked but it took forever.

[Rick]

Lets say I define a function (Call it delta) as:



So for all the reals we can show that:



Ie. Delta(1) = e, Delta(2) = e^2, Delta(-1) = 1/e

So what about:



Can we evaluate the function to -1?

[pvnk]

What is the probability that you will return from your ban?

[silur]

OH MY GOD WE HAVE A MATH DEDICATED THREAD *-*