Question

How do you solve `cosxsinx=cosx` in the interval `0<=x<=2pi`?

Answer 1

`x=pi/2" or "x=(3pi)/2`

Explanation:

`"rearrange and equate to zero"`

`cosxsinx-cosx=0`

`"take out common factor "cosx`

`cosx(sinx-1)=0`

`"equate each factor to zero and solve for x"`

`cosx=0rArrx=pi/2,(3pi)/2`

`sinx-1=0rArrsinx=1rArrx=pi/2`

`"Put the solutions together"`

`rArrx=pi/2" or "x=(3pi)/2tox in[0,2pi]`

Answer 2

`x=pi/2,(3pi)/2.`

Explanation:

Here, `cosxsinx=cosx=>cosxsinx-cosx=0`
`=>cosx(sinx-1)=0=>cosx=0orsinx-1=0`
`cosx=0or sinx=1`
`color(red)(=>x=(2k+1)pi/2,kinZorx=(4k+1)pi/2,kinZ)`
Now, `x in[0,2pi]=>x=pi/2,(3pi)/2.`