Question

Find the number of values of x which satisfy the equation

Answer 1

\[3\pi(1-\cos x)=2x\]

Answer 2

\[\frac{ 1-\cos x }{ x }=\frac{ 2 }{ 3\pi }\]
looks like sme limit x goes to 0

Answer 3

How about \( x= \frac {3 \pi}2\)

Answer 4

\[1-\cos 3\pi/2=1-\frac{ 1 }{ \sqrt{2} }\]

Answer 5

shud be 1+1/sqrt2

Answer 6

\(\cos( \frac {3 \pi }2 )=0\)

Answer 7

yes it works

Answer 8

This is the only one.
http://www.wolframalpha.com/input/?i=plot+%281-cos+x%29%2Fx

Answer 9

how do we knw if there are others

Answer 10

but this exam does not allow calculator,or wolframulators

Answer 11

Using L'Hospital's rule you can see that the limit of the quation goes to zero as x goes to zero.
Also you can show that the limit when x goes to infinity of the ratio is zero.
You can now finish it.

Answer 12

NOTE the limit part i just made up its not part of the question

Answer 13

so\[x=0,3\pi/2\]

Answer 14

\[3\pi(1-\cos0)=2(0)\]

Answer 15

0=0

Answer 16

thanks @eliassaab

Answer 17

looks like there are 3 solutions plot 3pi(1-cos x)=2x

Answer 18

x=0,5

Answer 19

http://www.wolframalpha.com/input/?i=plot+3pi%281-cos+x%29%3D2x

Answer 20

@eliassaab

Answer 21

5 roots actually
http://www.wolframalpha.com/input/?i=solve+3pi%281-cos+x%29+%3D+2x