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Hello, can anybody solve this equation: X - Sin(X) = π(pi)/2
Please explain the solution fully!
Please explain the solution fully!
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Maths Query!
- Thread starter Zishi
- Start date
Arranging the equation, u can get x= sin(x) + (pi/2)
sketch these graphs, u will get an idea where they intersect
Then iterate , u will get a root!!
or you can arrange in the form; sin x= x - (pi/2)
then x(n+1) =sin-1( X
- (pi/2))
iterate till u get the root!!
Reply if u got it
sketch these graphs, u will get an idea where they intersect
Then iterate , u will get a root!!
or you can arrange in the form; sin x= x - (pi/2)
then x(n+1) =sin-1( X
- (pi/2))iterate till u get the root!!
Reply if u got it
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I know about the graphical way, but how can we exactly find the roots? There should be any other algebric method to solve the equation? =s And sorry, I didn't understand your second way of solving this? =sazagen said:Arranging the equation, u can get x= sin(x) + (pi/2)
sketch these graphs, u will get an idea where they intersect
Then iterate , u will get a root!!
or you can arrange in the form; sin x= x - (pi/2)
then x(n+1) =sin-1( X
iterate till u get the root!!
Reply if u got it
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the easy way of solving this eqn using iteration (iteration is not included in AS but in A2). there's another way which is also not included in AS. However, i am not sure whether there is any algebraic method to solve it. Can't think of any......
hey, the graphical way is just to get an idea about the approximate root.
we use that approximate to get our iteration faster.
Suppose the actual root is 1.23456789
By sketching you can deduce that the root lies between 1 and 2. Then you just plug any value between 1 and 2 to iterate.
Iteration can give you the root to any decimal places you want but it ten becomes time consuming!!
we use that approximate to get our iteration faster.
Suppose the actual root is 1.23456789
By sketching you can deduce that the root lies between 1 and 2. Then you just plug any value between 1 and 2 to iterate.
Iteration can give you the root to any decimal places you want but it ten becomes time consuming!!
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Oh, well! I'm in AS Level and I don't know about 'Iteration', even I studied Additional Maths of O levels. And the questions asks to give the answer to 2 decimal places, so it means that approximation method may not be applied. So any other way, please?
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azagen said:
No no, I am talking about selecting two values of x and checking the signs of the answer.. as the book says 'Decimal search'.
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So is there no other method except iteration and drawing graph? :/
If iteration is in A2, then why this question has been asked in P1?
If iteration is in A2, then why this question has been asked in P1?
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P1? :?:
then bro, give the provide the complete question, if its from past papers then mention the question number and year as well...
then bro, give the provide the complete question, if its from past papers then mention the question number and year as well...
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Well, the question did not specifically asked u to find to theta.
I think u can also do it like this way,
rearrange which gives X - sin X - pi/2=0
let f(X)= X - sin X - pi/2
when X = 2.31,
f(2.31)= 1.9839 x 10^-4
which is approximately zero
thus x = 2.31
I think u can also do it like this way,
rearrange which gives X - sin X - pi/2=0
let f(X)= X - sin X - pi/2
when X = 2.31,
f(2.31)= 1.9839 x 10^-4
which is approximately zero
thus x = 2.31
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But from the wording correct to 2 decimal places, doesn't it seen that the questions wants us to solve the equation? =s
Anyways, thanks!
Anyways, thanks!
If I am not mistaken, this question is extracted from the endorsed A level mathematics textbook which does contain some material which has never been asked in any of the recent past papers..(2003 onward)
So, you can just skip such questions and cope with those questions what closely match the syllabus content..
So, you can just skip such questions and cope with those questions what closely match the syllabus content..
As far as I know, verifying a solution means that you just plug the value and if RHS=LHS then you can say it is verified.
In A2, we are often given a quadratic equation with no real roots
for example: 3(X^2)-x+7=0 and it asks us to verify that for e.g 2i-3 is a root, then we find (2i-3)^2 and we substitute in the left hand side(LHS) and if it reduces to zero, that is RHS=LHS , then it is verified that 2i-3 is a root.
In A2, we are often given a quadratic equation with no real roots
for example: 3(X^2)-x+7=0 and it asks us to verify that for e.g 2i-3 is a root, then we find (2i-3)^2 and we substitute in the left hand side(LHS) and if it reduces to zero, that is RHS=LHS , then it is verified that 2i-3 is a root.