Solving for x in a Trigonometric Equation

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Homework Statement



Sin x / x = 1/ (2)^1/2

how do I solve for x? I know x is 1.392

The Attempt at a Solution



I thought I might be able to use some series expansion, but it gets quite messy. Help please?
 
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Hi Jenkz! :smile:

(have a square-root: √ :wink:)

yup, only way is a series expansion or look it up in sinc tables …

see http://en.wikipedia.org/wiki/Sinc_function" :wink:
 
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I'm still not really getting anywhere with the answer :s

so far all I've got is:
sinx / x = 1/ √ 2 = 1 - X²/6 + X^4 / 120 - ...

Truncating the series and some re-arranging I get
1.752 = X² (1- X² /20 )

But I end upwith something completely wrong... :/
 
Ooo nvm. Thanks!
 
Jenkz said:
1.752 = X² (1- X² /20 )

But I end upwith something completely wrong... :/

Looks roughly correct to me (X = 1.4). :confused:
 
Yah.. I just did something silly. All works now! :D
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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