the answer is 0.24 (working in radians) how to you get this thanks!!
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Jan 7, 2007 #2 Integral Staff Emeritus Science Advisor Gold Member 7,185 55 There is no closed form solution. You might research something like a Newton's Method root finder. I did a quick fixed point interation and arrived at 0.24532 but it took over a 100 iterations to reach it. I started with a guess of .5 and iterated [tex] x = \frac { \sin x } {.99} [/tex]
There is no closed form solution. You might research something like a Newton's Method root finder. I did a quick fixed point interation and arrived at 0.24532 but it took over a 100 iterations to reach it. I started with a guess of .5 and iterated [tex] x = \frac { \sin x } {.99} [/tex]
Jan 7, 2007 #3 waht 1,464 3 I did an approximation by hand, Take the first terms of taylor series for sine [tex] sin(x) = x - \frac { x^3 } {6} + \frac {x^5} {120} [/tex] [tex] \frac { \sin x } {x} = 1 - \frac { x^2 } {6} + \frac {x^4} {120} = 0.99 [/tex] [tex] x^4 - 20x^2 + 1.2 = 0 [/tex] Substitute, a = x^2 [tex] a^2 - 20a + 1.2 = 0 [/tex] It's an easy quadratic equation, also got 0.2459674, and other solution 4.465366 which doesn't work. Last edited: Jan 7, 2007
I did an approximation by hand, Take the first terms of taylor series for sine [tex] sin(x) = x - \frac { x^3 } {6} + \frac {x^5} {120} [/tex] [tex] \frac { \sin x } {x} = 1 - \frac { x^2 } {6} + \frac {x^4} {120} = 0.99 [/tex] [tex] x^4 - 20x^2 + 1.2 = 0 [/tex] Substitute, a = x^2 [tex] a^2 - 20a + 1.2 = 0 [/tex] It's an easy quadratic equation, also got 0.2459674, and other solution 4.465366 which doesn't work.
Jan 8, 2007 #5 berkeman Mentor 55,874 5,939 willyf1 said: I looked for docs that referring to this problem but couldn't find any of those, do you have a clue? Click to expand... What about waht's solution?
willyf1 said: I looked for docs that referring to this problem but couldn't find any of those, do you have a clue? Click to expand... What about waht's solution?