Question

I need the to find smallest positive solution \[(\sin \pi x)(cosx-2)=0\]
\[\sin \pi x=0\] or \[cosx - 2=0\]
so equation of \[cosx - 2=0\] doesn't have a solution,does it?

Answer 1

Correct, cos(x) - 2 = 0 doesn't have a solution. Now if the sine of your particular angle pi*x = 0, then what must the angle be?

Answer 2

or do i need to write like
\[cosx-2=0\]
\[x-2=\pm \arccos a +2pik\]

Answer 3

0 isn't a positive solution...

Answer 4

pi*x=0|:pi
x=0?

Answer 5

@Kamille Did you mean
\[\large \cos(x-2)=0\] instead?

Answer 6

Because that wouldn't work.

Answer 7

iknow i need to sinx=0
x=pik

Answer 8

no, I meant cosx−2=0

Answer 9

If sin(pi*x) = 0, then pi*x must equal 0, pi, or some multiple of pi. Hence, if you want a positive solution, x = 1. Any other solution will either be negative, zero, or greater than 1.

Answer 10

oh, okay, thank you!