Question

How to solve sin5x = -sin3x?

Answer 1

`x={(2k+1)pi/2,kinZZ} U {(kpi)/4,kinZZ}`

Explanation:

We know that,

`color(red)((1)sinC+sinD=2sin((C+D)/2)cos((C-D)/2)`

`color(blue)((2)sintheta=0=>theta=kpi,kinZZ`

`color(blue)((3)costheta=0=>theta=(2k+1)pi/2,kinZZ`

Here,

`sin5x=-sin3x`

`=>sin5x+sin3x=0...tocolor(red)(Apply(1)`

`=>2sin((5x+3x)/2)cos((5x-3x)/2)=0`

`=>sin4xcosx=0`

`=>sin4x=0 or cosx=0`

`(i)sin4x=0=>4x=kpi,kinZZ...tocolor(blue)(Apply(2)`

`color(white)(..................)=>x=(kpi)/4,kinZZ`

`(ii)cosx=0=>x=(2k+1)pi/2,kinZZ...tocolor(blue)(Apply(3)`

Hence, from `(i) and(ii)`

`x={(kpi)/4,kinZZ}U{(2k+1)pi/2,kinZZ}`

Answer 2

`x in {1/4kpi:kinZZ} uu{(2k+1)/2pi:kinZZ}`

Explanation:

We have the following properties of `sin` which we will use to answer the question:

• `-sinx=sin(-x)`
• `sinx=sin(pi-x)`

Let `k` be an integer.

`sin5x=-sin3x=sin(-3x)=sin(pi+3x)`

`5x=-3x+2kpi` or `5x=pi+3x+2kpi`

so `8x=2kpi` or `2x=pi+2kpi`

so `x=1/4kpi` or `x=1/2pi+kpi=(2k+1)/2pi`

so `x in {1/4kpi:kinZZ} uu{(2k+1)/2pi:kinZZ}`