Question

How do you solve `sin2xsinx-cosx=0`?

Answer 1

The values of `x in { pi/2,(3pi)/2,pi/4,(3pi)/4,(5pi)/4,(7pi)/4}`

Explanation:

`Sin2x=2sinxcosx`
`sin2xcosx-cosx=2sinxcosxsinx-cosx=cosx(2sin^2x-1)`
`cosx=0` and `2sin^2x-1=0` `=>` `sinx=+-1/sqrt2`
Let the solutions `x in[0,2pi]`
The solution of the first equation is `(pi/2,(3pi)/2)`
The solution of the second equations are `(pi/4,(3pi)/4,(5pi)/4,(7pi)/4)`

Answer 2

`x=(kpi)/2 or x=(kpi)/4 k in ZZ`

Explanation:

Solving the given trigonometric equation `sin2xsinx - cosx` is determined by:

Substituting some trigonometric identities.

`color(blue)(sin2x=2sinxcosx)`

`color(red)(cos2x=1 - 2sin^2)`

Performing some calculations.

Factorizing `sin2x - cosx`

`" "`

Solving the given equation:

`sin2xsinx-cosx=0`

`rArr(color(blue)(2sinxcosx))sinx-cosx=0`

`rArr2sin^2xcosx-cosx=0`

`rArrcosx(color(red)(2sin^2x-1))=0`

`rArrcosx(color(red)(-cos2x))=0`

`rArr-cosxcos2x=0`

`cosx =0 rArr x=(kpi)/2" " k in ZZ`

Or

`cos2x=0`
`rArr 2x=(kpi)/2`
`rArr x=(kpi)/4 " " k in ZZ`

Therefore,

`x=(kpi)/2 or x=(kpi)/4 k in ZZ`