# Show that equation has no solution in $(0,2\pi)$

Hi I want to show that the equation $2=2 \cos(x)+x \sin(x)$ has no solution in $(0,2 \pi)$. Since it is algebraically impossible to solve this equation for $x$ I wanted to ask you whether one of you has an idea how to show this

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Let $f(x) = 2 \cos x + x \sin x$. Then $f'(x) = - 2 \sin x + \sin x + x \cos x = x \cos x - \sin x$ and $f''(x) = - x \sin x + \cos x - \cos x = - x \sin x$.
1. What is $f(0)$ and $f(2 \pi)$?
2. When $f''$ is negative (positive)?
3. What can you tell about the shape of the graph of $f'$? ($x \in (0, 2\pi)$)
4. What can you tell about the shape of the graph of $f$? ($x \in (0, 2\pi)$)
HINT: Note that for $\theta =x/2$, $$2x\sin \theta\cos\theta = x\sin x=2(1-\cos x)=4\sin^2\theta$$ Hence $\tan\theta=\theta$. Now look at the graphs of $y=\tan x$ and $y=x$ and try to show that they intersect only at $(0,0).$