How do you find the x-intercepts of $\frac{1}{2} x - sin (x) = 0$ $1/2 x - sin(x) = 0$ ?

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Answer 1 · 2018-07-17T16:56:43

$0 and \pm 1.8955$ $0 and +- 1.8955$ ., for 5 significant digits (5-sd)

Solutions of 1/2x - sin x = 0 are

x-intercepts of $y = x - 2 sin x$ $y = x - 2 sin x$ .

Obvious solution is x = 0. The other two located near

$\pm 2$ $+- 2$ by the first graph are transcendental.

Numerical iterative methods could give very high precision

approximations.Trial-and-error root-bracketing graphical method

gives 5-sd values.

The 1st graph locates two transcendental roots near $\pm 2$ $+- 2$ .

graph{y-x/2+sin x=0}
.
Simple or faster Newton-Raphson numerical iterative methods

would generate about (at the least )17-sd solutions, in long

precision. The starters are $x_{0} = +_{2}$ $x_0 = +_ 2$ , respectively..

Here, root-bracketing graphical method gives 5-sd solutions.

The graphs below gives 5-sd x-intercept $\pm 1.8955$ $+-1.8955$ .

Of course, they are not equidistant from O.

graph{y-x+2 sin x = 0[-1.896 -1.895 -.001 .001]}

graph{y-x+2 sin x = 0[1.8954 1.8956 -.001 .001]}

How do you find the x-intercepts of 1/2 x - sin(x) = 012x−sin(x)=01/2 x - sin(x) = 0?