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You can solve this to the accuracy of your liking by using Newton's method:

xn+1 = xn - f(xn) / f'(xn)

In this case, we'll say f(x) = x2 - cos(x)

f'(x) would then be 2x + sin(x)

Let's take a rough guess, and start with x0 = 0.5

x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536

x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916

x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712

x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229

x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297

x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296

Now we can test our answer:

0.824132312302522422962 = 0.67919406818110235182

cos(0.82413231230252242296) = 0.67919406818110235183

So we're accurate to the nearest ten quintillionth.

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13y ago
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Q: How do you solve x2 equals cos x?
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