Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

up vote 2 down vote accepted

This might help you.

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 $f$? ($x \in (0, 2\pi)$)
share|improve this answer

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 $(0,0).$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.