How to solve sin5x = -sin3x?

2 Answers
Apr 23, 2018

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}

Apr 23, 2018

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}