I have the equation $x^2 + x \cos(x) = 1 + \sin(x)$
and I need to prove that it has exactly two solutions.
What I used to do before when I had to prove an equation has one solution was:
used the intermediate value theorem to show there exists a solution and then used contradiction with Rolle's theorem to show that there's a unique one. But here I'm not so sure what to do.
Thanks for the help!
Best Answer
Let $f(x) = x^2 + x\cos(x) - \sin(x) - 1$. Then, $f'(x) = 2x - x \sin(x) = x(2 - \sin(x))$.
Now, $2 - \sin (x)$ is always positive, so the sign of $f'(x)$ is just the sign of $x$. In particular, the only root of $f'(x)$ is $0$.
Between any two roots of $f(x)$, there must be a point of zero derivative, by mean value theorem.
If there are more than two roots, then $f'(x)$ would have more than one root, contradiction.
Carefully apply intermediate value theorem and you shall be able to prove that two roots actually exist.