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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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2  
    
@HansLundmark I need to present some solution not just giving some numbers from some computer. – gen Jan 10 '13 at 15:33
    
Hint: Do a Taylor series expansion of $\cos x$ (say, up to order 6) on the left and then solve for x in the remaining equation. Now, explain this approach and figure out how to select the right value and what the error is. Regards – Amzoti Jan 10 '13 at 15:49
    
try using the fact that $ |\cos x| \le 1 $ and $ \cos (-x) = \cos (x)$ – Santosh Linkha Jan 10 '13 at 15:49
    
@gen: The word "solution" is ambiguous. It can mean both the value of $x$ which satisfies the equation, and the procedure for finding that value. When you say "an approx. solution", I think most people will interpret this as just the number, so if you want the procedure you should perhaps have said so in the question. – Hans Lundmark Jan 10 '13 at 18:30
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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