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I would like to get an approx. solution to the equation: $ \cos x = 2x$, I don't need an exact solution just some approx. And I need the solution without higher mathematics (without derive and things like that :) ).

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@nábob: Welcome to MSE! What have tried? Is this homework? If so, please tag it as such. What operations are you allowed to use if not derivatives? Regards –  Amzoti Jan 10 '13 at 15:39
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Deleting a question when a reasonable answer has been given can be viewed as an attempt to hide one's use of the site during an exam or contest. In any case, the question and answer might be useful to other users of the site. –  robjohn Jan 10 '13 at 16:15
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@gen: Deleting the question is not going to get answers that are easier to understand. If the answers are too advanced, edit the question or comment to tell what tools you have available or what level of book or class the problem came from. –  robjohn Jan 10 '13 at 16:55
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@robjohn Thank you next time I will do so, I am new here :). –  gen Jan 10 '13 at 17:02

3 Answers 3

up vote 15 down vote accepted

Take a pocket calculator, start with $0$ and repeatedly type [cos], [$\div$], [2], [=]. This will more or less quickly converge to a value $x$ such that $\frac{\cos x}2=x$, just what you want.

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Thank you it will be fine :) –  gen Jan 10 '13 at 16:44
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Or [$\cos$], [$2$], [$/$] if you are using an RPN calculator. –  robjohn Jan 10 '13 at 16:48
    
@robjohn just imagine if this question was deleted!. –  Neo Jan 10 '13 at 17:20
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Make sure you've got it set on RADIANS, not DEGREES. –  Michael Hardy Jan 10 '13 at 17:51

if $\cos x=2x$ which means $$x=\frac {1}{2.22131587} \,\mathrm{rad},$$

$$x=28.65957881\,\mathrm{Grad},$$

$$x=25.79362093\,\mathrm{Degree}$$

point: $$\cos x=1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!}+\cdots$$

$0 < x < +\frac {1}{2.5}\,\mathrm{rad}$ because $\cos x=2x$


This is also a mathematical method, i mean Estimate the amount through math without using a computer or calculator!.

$$\cos x \simeq 1-\frac {x^2}{2!}=2x,$$

$$x^2+4x-2=0,$$

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

where the $a=1$ and $b=4$ and $c=-2$ then $x\simeq 0.45$

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I need some reasing... –  gen Jan 10 '13 at 15:47
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What method did you use to get the approximate solution? –  robjohn Jan 10 '13 at 16:16
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I've written a complex computer program for solving these equations. –  Neo Jan 10 '13 at 16:24
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How is this supposed to help? The two upvoters might want to explain, as well. –  Did Jan 10 '13 at 16:26
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Neo: Sorry? What is based on what? –  Did Jan 10 '13 at 20:34

At small angles $\alpha ≈ sin\alpha$, so at the equation we can write $2 sin\alpha≈ cos \alpha$. From that $tan \alpha ≈ 1/2$. The solution to the last equation is less than a degree off the exact solution to the original equation.

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