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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.

Q: How do you solve x2 equals cos x?

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f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)

x=18

144X = X2 144X - X2 = 0 (144 - X)X = 0 X = 0 or X = 144

cos(x^2)=cos(x times x)

x = +13x = - 13

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f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)

x: x2 - 81 = 0

x=18

Divide both sides of the equation by x: x2 = 9x x2 / x = 9x / x x = 9

144X = X2 144X - X2 = 0 (144 - X)X = 0 X = 0 or X = 144

cos(x^2)=cos(x times x)

cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...

x = +13x = - 13

You cannot, in general, solve one equation with two unknown variables. x - y = x - x2 Subtract x from both sides: - y = - x2 Change signs: y = x2 And that is as far as you can go.

x(x - 1) = x2 + x x2 - x = x2 + x The only solution to this equation is x = 0

To solve for x: x2 = 11x - 10 x2 - 11x = -10 x2 - 11x + 10 =0 (x - 1)(x - 10) = 0 x = {1, 10)

x2 + 13x = -30 ∴ x2 + 13x + 30 = 0 ∴ (x + 3)(x + 10) = 0 ∴ x ∈ {-3, -10}