Show that the equation 2x^2=xsinx+cosx has exactly two real roots.

Question

Title is the question.. please help me

Answer 1

Let function f(x) = 2x^2 - x*sinx - cos x

f(1) = 2*1^2 - 1*sin 1 - cos 1

= 2*1 - sin 1 - cos 1

= 2 - sin1 - cos1 >0

f(1/2) = 2(1/2)^2 - (1/2)*sin(1/2) - cos(1/2)

= 2(1/4) - (1/2)*sin(1/2) - cos(1/2)

= 1/2 - 1/2*sin(1/2) - cos(1/2)

<0

So there is a root in the interval (-1,-1/2)

LIkewise, there is a root between (1/2,1) because

f(1/2) <0 and f(1) > 0

The intermediate value theorem says there is an x in the interval

where the function is zero, hence a solution.

The following table summarizes the results:

Finally, the function is PREDOMINANTLY quadratic, since the trig functions are bound by -1 and 1, there

are two solutions.

Answer 2

Consider the function f(x) = 2x^2 - xsin(x) - cos(x).

We want to show this function has exactly two zeros. First we show there can be at most 2. Looking at the derivative,

f'(x) = 4x - sin(x) - xcos(x) + sin(x) = 4x - xcos(x) = x(4 - cos(x))

Since cos(x) is at most 1, we conclude that f'(x) = 0 only when x = 0.

Moreover, f(x) is continuous and differentiable everywhere, so we can apply Rolle's to conclude that there

can be at most two zeros; if there were three zeros x1, x2, x3, then by Rolle's, there exists a1 and a2 such

that x1 < a1 < x2 < a2 < x3 and such that

f'(a1) = f'(a2) = 0,

a contradiction since we concluded f'(x) = 0 only when x = 0 (so at most one x value).

Finally, there are at least two zeros, since

f(-pi) = 2(-pi)^2 - (-pi)sin(-pi) - cos(-pi) = 2pi^2 + 1

f(0) = -1

f(pi) = 2(pi)^2 - (pi)sin(pi) - cos(pi) = 2pi^2 + 1

and by continuity of f(x), there must exist a zero between -pi and 0, and a zero between 0 and pi.

So the number of zeros is at most 2, and at least 2; so there are exactly 2 zeros. Hence there are exactly

two values of x satisfying

2x^2 = xsin(x) + cos(x)