how do you solve 0=2cosx-x
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An approximation to the solution can be obtained by writing cos(x)=1-½x2 which is an expansion of the series for cos(x) up to the first two terms. So 0=2-x2-x or x2+x-2=0=(x+2)(x-1), therefore x=-2 or 1 (radians) approximately. In fact, there's only one solution and it happens to be close to 1, as we'll see below.

It can be solved graphically by plotting y=2cos(x)-x and measuring the x-intercept(s).

Newton's Method is another way, but first we find the derivative y'=-2sin(x)-1.

xn+1=xn-y(xn)/y'(xn) is an iterative process and assumes that the iterations of x will converge. To start the process we have to choose an arbitrary x0. Ideally this value should be close to the solution. y(0)=2-0=2 and y(π/3)=y(1.0472)=1-1.0472=-0.0472 (very close to zero). So there's a sign change between x=0 and π/3 radians (that is, 60°). We could therefore choose x0=π/3 or 0. Let x0=π/3.

x1=π/3-y(π/3)/y'(π/3)=1.02992..., x2=1.02986653=x3. This is therefore a solution to the given equation. A rough graph confirms that there is only one solution: x=1.0299 approx.

ago by Top Rated User (1.2m points)

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